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Before studying this page, make sure you understand the difference between the slope and tangent. If you need to refresh your memory on equations of lines, you can find a quick review here.
One thing I review in my class is knowing the difference between the slope and tangent. When you have a question that asks for one of these, read the question again carefully.
Slope
If the question is asking for the slope (or the slope of the tangent line), you just need to find m and your answer will be the derivative, or if they ask for the slope at a specific point, your answer will be number.
Tangent
If the question is asking for a tangent, they are asking you to find the equation of the tangent line. To do this, you need to find the slope first (by taking the derivative and evaluating it at the point), then plug the slope and the point into one of these line equations:
\(
\begin{array}{rclcl}
(y - y_0) & = & m(x - x_0) & --- & \text{Point-Slope Form} \\
y & = & mx + b & --- & \text{Slope-Intercept Form}
\end{array}
\)
Your final answer in this case is a linear equation. If you are in my class, I require your answer to be in slope-intercept form. Check the question carefully to see what form your teacher asks for.
Another twist to this is that not all teachers are consistent. Sometimes they will say 'tangent' and mean 'slope'. You will need to get to know how your teacher uses the term or ask them to clarify what they mean. However, if they say 'tangent line' (without saying 'the slope of'), this always means the equation of the tangent line.
Remember to read the question carefully and give the answer asked for, in the proper form. In my class, if you give the tangent line when I ask for the slope, you lose points and vice-versa. Hopefully, your teacher will help you learn the difference by doing this also.
Slope |
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Remember from the main derivative page that we said whenever you think about derivative you should automatically think of slope and vice-versa. If you do that then, when you look at the title of this section, your first thought should be derivative.
And that's all there is to it. To find the slope, just take the derivative. The derivative is just the slope at every point. If you are given a specific point, you just plug that into the derivative and the result is the slope at that point.
Let's work a few practice problems before going on. For these problems, calculate the slope of the function at the point, if given.
\(y=4-9x^2\)
slope \((6,-320)\)
Problem Statement |
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\(y=4-9x^2\)
slope \((6,-320)\)
Final Answer |
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The slope of the tangent line to the graph \(y=4-9x^2\) through the point \((6,-320)\) is \(m=-108\). |
Problem Statement |
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\(y=4-9x^2\)
slope \((6,-320)\)
Solution |
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First, we check that the point is on the graph of the function.
\( y = 4 - 9 (6^2) = 4 - 9(36) = 4 - 324 = -320 \)
So the point \( (6,-320) \) is on the graph of \( y = 4-9x^2 \).
So we have one point on the graph. We need to find the slope at that point. To do that, we need the derivative.
\( y'(x) = 0 - 9(2x) = -18x \)
At the point \( (6,-320) \), the slope is \( m = y'(2) = -18(6) = -108 \).
Final Answer |
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The slope of the tangent line to the graph \(y=4-9x^2\) through the point \((6,-320)\) is \(m=-108\). |
close solution |
Tangent Line |
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To find the equation of a tangent line, there are two main steps.
Step 1: Calculate the slope.
Step 2: Use the slope from step one and a point (usually given) and find the equation of the tangent line.
Note: This procedure works only if the given point is on the graph of the function. [See here if the point is not on the graph.]
Let's go through an example to demonstrate this process.
Example
Find the equation of the tangent line to the curve
\( f(x)=x^3/3 + x^2 + 5 \) at the point \((3,23)\).
First, let's look at a graph of this function and get an idea of what the problem is asking for. This is the graph of \(f(x)\) (blue line) and the point \( (3,23)\) is shown with an arrow. The idea is that we need to take the derivative of the function \(f(x)\) plug in the point \((3,23)\) into the derivative. Since the derivative gives the equation of the slope of \(f(x)\), plugging in the point \((3,23)\) gives us the slope at that specific point. So, let's get started.
First, we need to make sure the point is on the graph. It looks like it is from the graph to the right but that is not good enough. It could be slightly off and not show up on this graph. To make sure, we evaluate \(f(3)\). If it is equal to \(23\), then we are good.
\( f(3) = 3^3/3 + 3^2 + 5 =\) \( 9 + 9 + 5 = 23 \)
So, this tells us that the point \((3,23)\) is on the curve \(f(x)\) and we can use the procedure outlined above.
Step 1: Calculate the slope.
Now, find the derivative of \(f(x)\).
\( f'(x) = (3x^2)/3 + 2x + 0 \)
In this step, we used the constant multiple rule, the power rule and constant rule.
Simplifying, we get \( f'(x) = x^2 + 2x \).
Just to clarify, this equation, \( f'(x)\) is the equation of the slope of \(f(x)\) at every point. In this problem, we want the slope \((3,23)\), so we let \( x = 3 \).
So we get the slope \( m = f'(3) = 3^2 +2(3) = 15 \).
Step 2: Find the equation of the tangent line.
Now we can use either the point-slope form of the equation of a line or the slope-intercept form. I will use the point-slope form, \( y-y_1=m(x-x_1) \).
\(\begin{array}{rcl}
y-23 & = & 15(x-3) \\
y & = & 15x - 45 + 23 \\
y & = & 15x - 22
\end{array}\)
Final Answer |
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The equation of the tangent line to the curve \( f(x)=x^3/3 + x^2 + 5 \) at the point \((3,23)\) is \( y=15x-22 \). |
To verify that our answer is correct, we can graph the function and the tangent line on the same set of axes, to make sure the line looks tangent to the curve and goes through the point. Here is what the result looks like.
Now, there are some things you need to watch for when you are asked to find a tangent line.
First, If you are given only the x-value, you can plug that value into the original function, \(f(x)\) to get \(y\).
A second twist is, given a point \((x_1,y_1)\), depending on how the question is worded, you are not guaranteed that the point is on the function. If it isn't, you need to do a whole different set of procedures (outlined in the panel below).
For example, you may get a question worded like this:
Find the equation of the tangent line to the graph \(g(x)\) through the point \((x_1,y_1)\).
In this case, the point \((x_1,y_1)\) may or may not be on the graph. To check it, we evaluate \(g(x_1)\). If \(g(x_1) = y_1\), we can use the procedure above. To be safe, it is always good to graph the function and evaluate the function at the x-value to make sure you get the y-value to determine if the given point is on the graph.
Finding the equation of the tangent line through a point that is NOT on the graph is an added twist to this problem. However, unlike most twists, this one takes quite a bit of thought to work through. So it is good to think about it before it comes up on an exam. First, let's discuss what the problem looks like. [ Hints: Do not read through this quickly. Stop and think about what this says at each step or you will not understand it. It may also help to draw a graph as you go through it. And you may need to read this several times to get your head around it. ]
The idea is that we have a function and we want to find a tangent line through a point that is not on the function. We can't just plug the x-value into the derivative to find the slope because the tangent line does not go through that x-value. We need to find the point on the function.
To demonstrate this procedure, we will use the same function as the previous example except we want the tangent line to go through the point \((4,30)\). To check that the point is not on the function, we plug \(x=4\) into the function to get
\( f(4) = ((4^3)/3+4^2+5 = 64/3+16+5 = 127/3 \neq 30 \)
This is verified by the graph on the right (never rely on the graph to give this information).
Here is what we can use from the previous example above.
The function is \( f(x)=x^3/3 + x^2 + 5 \).
The derivative is \( f'(x) = x^2 + 2x \).
We are not given information about where the tangent line touches the function. So we will call this unknown point \((a,b)\). We know the slope at this point \( m = a^2+2a \) and the function value at this point is \( b = (a^3)/3+a^2+5 \) (since the point \((a,b)\) is on the graph of the function). We also know that the point given in the problem statement \( (4,30) \) as well as the unknown point \((a,b)\) are on the tangent line. Using the point-slope form of the equation of the line \(y-y_1 = (x-x_1)m \) we can put all the information into one equation.
\(
\begin{array}{rcl}
y-y_1 & = & (x-x_1)m \\
b-30 & = & (a-4)(a^2+2a) \\
(a^3)/3+a^2+5 - 30 & = & a^3 + 2a^2 - 4a^2 - 8a \\
(a^3)/3+a^2-25 - a^3 +2a^2+8a = 0 \\
(-2a^3)/3 + 3a^2 + 8a-25 & = & 0
\end{array}
\)
This last equation is not factorable, so we use the root finding capability on our calculator or graphing utility to solve for \(a\). We know from algebra that there is either one real root or three real roots. We determine that there are three real roots whose values are
\(
\begin{array}{rcr}
a_1 & = & -3.13678 \\
a_2 & = & 2.19815 \\
a_3 & = & 5.43863
\end{array}
\)
So which one do we choose? We were not given information in the problem statement about which one, so we can't eliminate any of them. So what do we do with all three? Since there are three different points where there is a tangent line through the point \( (4,30)\), there are actually three tangent lines as well. So we need to find all three equations.
The details involve only algebra now, so we will present the answers and the graph.
point | tangent line | |
\( (-3.13678, 4.55495 )\) | \( y = 3.56582x+15.73661 \) | |
\(( 2.19815, 13.37225 )\) | \( y = 9.22816x-6.912637 \) | |
\( (5.43863, 88.20122)\) | \(y = 40.455956x-131.82375 \) |
Note: We solved this analytically, i.e. not by guessing from the graph. Most instructors will want you to work it this way. Check with your instructor to see what they require.
Okay, before we go on, let's work some practice problems. Unless otherwise instructed, calculate the tangent line of the function/equation through the given point. Don't forget to check that the point is on the graph. Give your answers in slope-intercept form.
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new. |
Derivative - Slope, Tangent, Normal - Practice Problems Conversion |
[A01-956] - [A02-957] - [A03-958] - [A04-960] - [A05-959] - [A06-961] - [A07-962] - [A08-963] - [A09-964] - [A10-965] |
[A11-966] - [A12-967] - [A13-968] - [A14-969] - [A15-970] - [A16-971] - [A17-972] - [A18-1314] - [A09-1911] - [B01-1918] |
Please update your notes to this new numbering system. The display of this conversion information is temporary. |
\(g(x)=x^3-3x+1\)
tangent \((1,-1)\)
Problem Statement |
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\(g(x)=x^3-3x+1\)
tangent \((1,-1)\)
Final Answer |
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The equation of the tangent line to \(g(x)=x^3-3x+1\) through the point \((1,-1)\) is \(y=-1\). |
Problem Statement |
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\(g(x)=x^3-3x+1\)
tangent \((1,-1)\)
Solution |
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First, we need to determine if the point \((1,-1)\) is on the graph of g(x). We could graph it or plug in the point \(x=1\) and see if we get \(y=-1\). Plugging in the point is more exact (since our graph may LOOK like the point is on the curve when it may not be). So let's evaluate \(g(x)\) at \(x=1\).
\(g(1) = 1^3 - 3(1)+1 = 1 - 3 + 1 = -1\)
So, the point \((1,-1)\) is on the curve.
Step 1: Calculate the derivative and slope.
\(g'(x) = 3x^2 - 3 \)
This is the equation that gives the slope of \(g(x)\) at every point on the curve. To get the slope at the specific point \(x=1\), we evaluate \(g'(1) = 3(1^2)-3 = 3-3 = 0 \).
Step 2:Using the slope from step 1 and the point given in the problem statement, find the equation of the tangent line.
Okay, so the slope \(m=0\), which means the point-slope form of the tangent line is:
\( y - (-1) = 0(x - 1) ~~~ \to ~~~ y = -1 \).
This equation is in slope-intercept form ( \(y=mx+b\) ), so we are done.
Just to make sure this is correct, here is the graph. Let's analyze this a bit. First, the blue line is the graph of \(g(x)\) and the red line is the tangent line. Notice that this is no ordinary tangent line. This is a horizontal tangent line (since \(m=0\)). This horizontal tangent line occurs at a low point (called a relative minimum). Another horizontal tangent line would occur at \(x=-1\), which is a peak (called a relative maximum) on the curve. You can learn more about this on the graphing page.
Final Answer |
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The equation of the tangent line to \(g(x)=x^3-3x+1\) through the point \((1,-1)\) is \(y=-1\). |
close solution |
\(\displaystyle{f(x)=\frac{2}{3}x^3+x^2-12x+6}\)
tangent \(x=0\)
Problem Statement |
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\(\displaystyle{f(x)=\frac{2}{3}x^3+x^2-12x+6}\)
tangent \(x=0\)
Final Answer |
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The equation of the tangent line to \(\displaystyle{f(x)=\frac{2}{3}x^3+x^2-12x+6}\) at \(x=0\) is \(y=-12x+6\). |
Problem Statement |
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\(\displaystyle{f(x)=\frac{2}{3}x^3+x^2-12x+6}\)
tangent \(x=0\)
Solution |
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Step 1: Find the derivative of \(f(x)\).
\( \displaystyle{ f'(x) = \frac{2}{3}(3x^2) + 2x -12 = 2x^2 +2x - 12 }\)
Now use the value \(x=0\), to calculate the slope.
\(f'(0) = 2(0^2) + 2(0)-12 = -12\)
Step 2: Find the equation of the tangent line.
We know that \(m=-12\) and \(x=0\) but, we need the y-value of the function, \(f(0) = 0 + 0 - 0 +6 = 6\). So our point is \((0,6)\). Now we can use the point-slope form of the equation of a line to get the tangent line.
\( y-6 = -12(x-0) ~~~ \to ~~ y = -12x+6\)
To verify the correctness of this solution, it's a good idea to graph both the function and the tangent line on the same set of axes.
Final Answer |
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The equation of the tangent line to \(\displaystyle{f(x)=\frac{2}{3}x^3+x^2-12x+6}\) at \(x=0\) is \(y=-12x+6\). |
close solution |
\(\displaystyle{h(x)=2x+\frac{1}{\sqrt{x}}}\)
tangent \((1,3)\)
Problem Statement |
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\(\displaystyle{h(x)=2x+\frac{1}{\sqrt{x}}}\)
tangent \((1,3)\)
Final Answer |
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The equation of the tangent line to the curve \(\displaystyle{h(x)=2x+\frac{1}{\sqrt{x}}}\) at the point \((1,3)\) is \(y=(3/2)x+3/2\). |
Problem Statement |
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\(\displaystyle{h(x)=2x+\frac{1}{\sqrt{x}}}\)
tangent \((1,3)\)
Solution |
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We need to verify that the point \((1,3)\) is on the graph of \(h(x)\).
\( h(1) = 2(1)+1/\sqrt{1} = 3 \)
Since \(h(1)=3\) we know that the point \((1,3)\) is on the curve.
Step 1: Find the derivative and calculate the slope.
First, we rewrite the square root term to be able to use the power rule.
\( h(x) = 2x + x^{-1/2} \)
In this step, we rewrote the square root as a \(1/2\) power and then moved the term to the numerator, which entailed changing the sign of the exponent. Now we can take the derivative.
\( h'(x) = 2 + (-1/2)x^{-3/2} \)
To find the slope at the point \((1,3)\), we need to calculate \( h'(1) \).
\( h'(1) = 2 + (-1/2)(1)^{-3/2} = 2 - 1/2 = 3/2 \)
Step 2: Find the equation of the tangent line.
We have the slope \(m=3/2\) and the point \((1,3)\).
Let's use the slope-intercept form this time.
\( y = mx+b \) |
\( 3 = (3/2)(1)+b \) |
\( b = 3 - 3/2 = 3/2 \) |
\( y = (3/2)x + 3/2 \) |
To check the answer, it would be a good idea to graph the function and the tangent line on the same set of axes.
Final Answer |
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The equation of the tangent line to the curve \(\displaystyle{h(x)=2x+\frac{1}{\sqrt{x}}}\) at the point \((1,3)\) is \(y=(3/2)x+3/2\). |
close solution |
\(\displaystyle{g(x)=\frac{2}{3}x^3+\frac{5}{2}x^2+2x+1}\)
tangent \(x=3\)
Problem Statement |
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\(\displaystyle{g(x)=\frac{2}{3}x^3+\frac{5}{2}x^2+2x+1}\)
tangent \(x=3\)
Final Answer |
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The equation of the tangent line to the function \(\displaystyle{ g(x)=\frac{2}{3}x^3+\frac{5}{2}x^2+2x+1 }\) at \(x=3\) is \(y=35x-115/2\). |
Problem Statement |
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\(\displaystyle{g(x)=\frac{2}{3}x^3+\frac{5}{2}x^2+2x+1}\)
tangent \(x=3\)
Solution |
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Step 1: Find the slope of the function at \(x=3\).
This means we find the derivative and evaluate it \(x=3\).
\(\displaystyle{ g'(x) = \frac{2}{3}(3x^2) + \frac{5}{2}(2x) + 2 = 2x^2 + 5x +2 }\)
\( g'(3) = 2(3^2) + 5(3) +2 = 2(9)+15 +2 = 18+15 +2= 35 \)
So the slope \( m = 35 \).
Step 2: Find the equation of the tangent line.
Let's use the point-slope form. We have the slope \( m = 35 \) and the x-value of the point \(x=3\). In order to find an equation of the line, we need a y-value. We can get that from \(g(x)\) since the tangent line and \(g(x)\) meet at the point of tangency. So, \( y = g(3) = (2/3)3^3 + (5/2)(3^2) + 2(3) + 1 = 18 + 45/2 + 6 + 1 = \)
\( 25 + 45/2 = 50/2 + 45/2 = 95/2 \) and the point is \((3, 95/2)\).
We can now find the equation of the line.
\( y - 95/2 = 35(x - 3) \) |
\( y = 35x - 105 + 95/2 \) |
\( y = 35x - 210/2 + 95/2 \) |
\( y = 35x -115/2 \) |
Final Answer |
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The equation of the tangent line to the function \(\displaystyle{ g(x)=\frac{2}{3}x^3+\frac{5}{2}x^2+2x+1 }\) at \(x=3\) is \(y=35x-115/2\). |
close solution |
\(g(x)=(x^2+1)(2-x)\)
tangent \((2,0)\)
Problem Statement |
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\(g(x)=(x^2+1)(2-x)\)
tangent \((2,0)\)
Final Answer |
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The equation of the tangent line to the curve \(g(x)=(x^2+1)(2-x)\) at the point \((2,0)\) is \(y=-5x+10\). |
Problem Statement |
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\(g(x)=(x^2+1)(2-x)\)
tangent \((2,0)\)
Solution |
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We need to check that the point \((2,0)\) is on the curve.
\( g(2) = (2^2+1)(2-2) = 5(0) = 0 \)
Since \( g(2) = 0 \), this tells us that the point \((2,0)\) is on the graph of \(g(x)\).
Step 1: Find the derivative and slope.
Using the product rule, we have
\( g'(x) = (x^2+1)(-1) + (2-x)(2x) = -x^2 - 1 + 4x - 2x^2 = -3x^2 + 4x - 1\)
The slope at \(x=2\) is \(g'(2) = -3(2^2) + 4(2) - 1 = -12 + 8 - 1 = -5\)
Step 2: Find the equation of the tangent line.
Using the slope-intercept form, we have \( m=-5 \) at the point \((2,0)\).
\( y=mx+b \) |
\( 0 = -5(2)+b \) |
\( b = 10 \) |
To check this answer, graph the function and the tangent line on the same set of axes.
Final Answer |
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The equation of the tangent line to the curve \(g(x)=(x^2+1)(2-x)\) at the point \((2,0)\) is \(y=-5x+10\). |
close solution |
\(y=1-9x^2\)
tangent \((2,-35)\)
Problem Statement |
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\(y=1-9x^2\)
tangent \((2,-35)\)
Final Answer |
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The equation of the tangent line to the graph \(y=1-9x^2\) through the point \((2,-35)\) is \(y=-36x+37\). |
Problem Statement |
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\(y=1-9x^2\)
tangent \((2,-35)\)
Solution |
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First, we need to determine if the given point is on the graph.
\( y = 1 - 9 (2^2) = 1 - 9(4) = 1-36 = -35 \)
So the point \( (2,-35) \) is on the graph of \( y = 1-9x^2 \).
So we have one point on the graph. We need to find the slope at that point. To do that, we need the derivative.
\( y' = 0 - 9(2x) = -18x \)
At the point \( (2,-35) \), the slope is \( m = y'(2) = -18(2) = -36 \).
So we have the slope \( m = -36 \) and one point \( (2,-35) \), so we can find the equation of the line.
\( y - (-35) = -36 (x-2) \) |
\( y = -36x +72 - 35 \) |
\( y = -36x + 37 \) |
It is a good idea to check the answer by graphing the function and the tangent line on the same set of axes.
Final Answer |
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The equation of the tangent line to the graph \(y=1-9x^2\) through the point \((2,-35)\) is \(y=-36x+37\). |
close solution |
Find the tangent line to the graph of \(\displaystyle{y=(2+x)e^{-x}}\) at the point \((0,2)\)
Problem Statement |
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Find the tangent line to the graph of \(\displaystyle{y=(2+x)e^{-x}}\) at the point \((0,2)\)
Solution |
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video by Krista King Math
close solution |
\(f(x)=(1+x)^2\)
tangent \(x=0\)
Problem Statement |
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\(f(x)=(1+x)^2\)
tangent \(x=0\)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{f(x)=(1-2x)^{3/2}}\)
tangent \(x=0\)
Problem Statement |
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\(\displaystyle{f(x)=(1-2x)^{3/2}}\)
tangent \(x=0\)
Solution |
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video by Krista King Math
close solution |
\(f(x)=\sin(x)\)
tangent \(x=0\)
Problem Statement |
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\(f(x)=\sin(x)\)
tangent \(x=0\)
Solution |
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video by Krista King Math
close solution |
\(f(x)=\cos(x)\)
tangent \(x=\pi/2\)
Problem Statement |
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\(f(x)=\cos(x)\)
tangent \(x=\pi/2\)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{f(x)=\frac{x^2-x+1}{3x+2}}\)
tangent \((0,1/2)\)
Problem Statement |
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\(\displaystyle{f(x)=\frac{x^2-x+1}{3x+2}}\)
tangent \((0,1/2)\)
Solution |
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video by PatrickJMT
close solution |
\(f(x)=(2x-1)^5(x+1)\)
tangent \(x=1\)
Problem Statement |
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\(f(x)=(2x-1)^5(x+1)\)
tangent \(x=1\)
Final Answer |
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\(y=21x-19\) |
Problem Statement |
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\(f(x)=(2x-1)^5(x+1)\)
tangent \(x=1\)
Solution |
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video by Krista King Math
Final Answer |
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\(y=21x-19\) |
close solution |
Find the horizontal and vertical tangent lines to the function \(f(x)=x\sqrt{1-x^2}\) on the interval \( -1 \leq x \leq 1 \).
Problem Statement |
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Find the horizontal and vertical tangent lines to the function \(f(x)=x\sqrt{1-x^2}\) on the interval \( -1 \leq x \leq 1 \).
Solution |
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video by Krista King Math
close solution |
If \(f(x)=7xe^x\), find the tangent line at \((0,0)\).
Problem Statement |
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If \(f(x)=7xe^x\), find the tangent line at \((0,0)\).
Solution |
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video by MathTV
close solution |
Linearization |
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Linearization is essentially the equation of the tangent line at a point. Here are a few practice problems to help you understand this.
Find the linearization of \(\displaystyle{f(x)=\frac{1}{\sqrt{7+x}}}\) at \(x=0\).
Problem Statement |
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Find the linearization of \(\displaystyle{f(x)=\frac{1}{\sqrt{7+x}}}\) at \(x=0\).
Solution |
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video by PatrickJMT
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Find the linearization of \(f(x)=\cos(x)\) at \(x=\pi/2\).
Problem Statement |
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Find the linearization of \(f(x)=\cos(x)\) at \(x=\pi/2\).
Solution |
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video by Krista King Math
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Normal Line |
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The normal line is perpendicular to the curve and, therefore, also perpendicular to the tangent line. To find the equation of the normal line at a point, follow the same procedure above, expect after finding the slope of the tangent line, take the negative reciprocal of the slope to get the slope of the normal line. Then use the point to the find the equation of normal line.
In the example above, the slope of the normal line is \( m=-1/15 \). Using this and the point \( (x_1, y_1) = (3,23)\), the equation of the normal line is
\(
\begin{array}{rcl}
y-y_1 & = & m(x-x_1) \\
y-23 & = & (-1/15)(x-3) \\
y & = & -x/15 + 1/5 + 23 \\
y & = & -x/15 + 116/5
\end{array}
\)
The graph of this normal line is shown in the plot on the right. It is shown as the green almost horizontal line. Notice that it doesn't look exactly perpendicular to the tangent line. This is because the scales on the axes are not equivalent. However, you know from the equations that they are perpendicular. The slope of the tangent line is \( m_t = 15 \) and the slope of the normal line is \( m_n = -1/15 \) and since \( m_n = -1/m_t \), they are perpendicular.
Calculate the equation of the normal line to the graph given by the equation, that goes through the given point. Again, you need to check to make sure that the point is on the graph. Give your answers in slope-intercept form.
\(y=x^2\)
normal \((-2,4)\)
Problem Statement |
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\(y=x^2\)
normal \((-2,4)\)
Solution |
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video by Krista King Math
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\(y=2x^2+3x-5\)
normal \((2,9)\)
Problem Statement |
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\(y=2x^2+3x-5\)
normal \((2,9)\)
Solution |
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video by Krista King Math
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\(y=5-x-2x^2\)
normal \((-1,4)\)
Problem Statement |
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\(y=5-x-2x^2\)
normal \((-1,4)\)
Solution |
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video by Krista King Math
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\(y=x^4+2e^x\)
normal \((0,2)\)
Problem Statement |
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\(y=x^4+2e^x\)
normal \((0,2)\)
Solution |
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video by Krista King Math
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\(y=(2+x)e^{-x}\)
normal \((0,2)\)
Problem Statement |
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\(y=(2+x)e^{-x}\)
normal \((0,2)\)
Solution |
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video by Krista King Math
close solution |