## 17Calculus - Calculus 1 - Practice Exam 1 (Semester B)

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This is the first exam for first semester single variable calculus.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2.5 hours

Questions

9

Total Points

125

Tools

Calculator

not allowed

Formula Sheet(s)

one page 8.5x11 or A4; both sides okay

Other Tools

none

Instructions:
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

[10 points] Answer either TRUE or FALSE for each of these statements.
(a) If $$f(x)$$ is undefined at $$x=c$$, then $$\displaystyle{ \lim_{x\to c}{f(x)} }$$ does not exist.
(b) If $$\displaystyle{ \lim_{x\to c}{g(x)} = 0 }$$, then there must exist a number k such that $$g(k) \lt 0.001$$.
(c) If $$f(c) = L$$, then $$\displaystyle{ \lim_{x\to c}{f(x)} = L }$$.
(d) If $$\displaystyle{ \lim_{x\to c}{f(x)} = L }$$, then $$f(c) = L$$.
(e) The function $$h(x)=|x-1|/(x-1)$$ is continuous on $$(-\infty, \infty)$$.
(f) If $$\displaystyle{ \lim_{x\to c}{f(x)} = L }$$, then $$f(x)$$ is continuous at $$x=c$$.
(g) For any polynomial function $$p(x)$$, $$\displaystyle{ \lim_{x\to a}{p(x)} = p(a) }$$.

Problem Statement

[10 points] Answer either TRUE or FALSE for each of these statements.
(a) If $$f(x)$$ is undefined at $$x=c$$, then $$\displaystyle{ \lim_{x\to c}{f(x)} }$$ does not exist.
(b) If $$\displaystyle{ \lim_{x\to c}{g(x)} = 0 }$$, then there must exist a number k such that $$g(k) \lt 0.001$$.
(c) If $$f(c) = L$$, then $$\displaystyle{ \lim_{x\to c}{f(x)} = L }$$.
(d) If $$\displaystyle{ \lim_{x\to c}{f(x)} = L }$$, then $$f(c) = L$$.
(e) The function $$h(x)=|x-1|/(x-1)$$ is continuous on $$(-\infty, \infty)$$.
(f) If $$\displaystyle{ \lim_{x\to c}{f(x)} = L }$$, then $$f(x)$$ is continuous at $$x=c$$.
(g) For any polynomial function $$p(x)$$, $$\displaystyle{ \lim_{x\to a}{p(x)} = p(a) }$$.

Solution

 (a) is FALSE. If there is a hole at $$x=c$$ then the limit exists even though the $$f(c)$$ is undefined. (b) is TRUE. If the limit exists, then no matter how close we get to $$x=c$$, there has to be a number that is less than 0.001 away from zero. If this was not TRUE, then there would be a gap and the limit would not exist. (c) is FALSE. We could set up a function where $$f(c) = L$$ but the limit goes to another number, $$\pm\infty$$ or does not exist. (d) is FALSE. This possibility exists since we could have a hole at $$f(c)$$ or $$f(c)$$ could be defined at a point other than L. (e) is FALSE. If you look at the plot, you will see that there is a jump at $$x=1$$ and therefore $$h(x)$$ is not continuous there. (f) is FALSE. This case does not hold when there is a hole at $$f(c)$$. (g) is TRUE. Polynomials are continuous for all real numbers, so the limit exists at all real numbers.

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[5 points] Evaluate $$\displaystyle{ \lim_{x \to -1}{ \sqrt{10-2x^3} } }$$.

Problem Statement

[5 points] Evaluate $$\displaystyle{ \lim_{x \to -1}{ \sqrt{10-2x^3} } }$$.

$$\displaystyle{ \lim_{x \to -1}{ \sqrt{10-2x^3} } = 2\sqrt{3}}$$

Problem Statement

[5 points] Evaluate $$\displaystyle{ \lim_{x \to -1}{ \sqrt{10-2x^3} } }$$.

Solution

 As usual, we try direct substitution first. $$\displaystyle{ \lim_{x \to -1}{ \sqrt{10-2x^3} } }$$ $$\displaystyle{ \sqrt{10-2(-1)^3} = \sqrt{12} = 2\sqrt{3} }$$

$$\displaystyle{ \lim_{x \to -1}{ \sqrt{10-2x^3} } = 2\sqrt{3}}$$

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[10 points] Calculate, using the idea of limits, the values of the constants b and c that will make the function $$h(x)$$ continuous everywhere.
$$\displaystyle{ h(x) = \left\{ \begin{array}{cl} x + 1 & 1 \lt x \lt 3 \\ x^2 + bx + c & |x-2| \geq 1 \end{array} \right. }$$

Problem Statement

[10 points] Calculate, using the idea of limits, the values of the constants b and c that will make the function $$h(x)$$ continuous everywhere.
$$\displaystyle{ h(x) = \left\{ \begin{array}{cl} x + 1 & 1 \lt x \lt 3 \\ x^2 + bx + c & |x-2| \geq 1 \end{array} \right. }$$

$$b = -3, c = 4$$

Problem Statement

[10 points] Calculate, using the idea of limits, the values of the constants b and c that will make the function $$h(x)$$ continuous everywhere.
$$\displaystyle{ h(x) = \left\{ \begin{array}{cl} x + 1 & 1 \lt x \lt 3 \\ x^2 + bx + c & |x-2| \geq 1 \end{array} \right. }$$

Solution

Since $$x+1$$ and $$x^2+bx+c$$ are polynomials, they are continuous on their respective domains. We just need to check the endpoints at $$x=1$$ and $$x=3$$.
To remove the absolute values, we have 2 cases.
Case 1: $$x-2 \geq 1 \to x \geq 3$$.
Case 2: $$-(x-2) \geq 1 \to -x+2 \geq 1 \to -x \geq -1 \to x \leq 1$$.
So $$h(x)$$ can be rewritten as
$$\displaystyle{ h(x) = \left\{ \begin{array}{cl} x^2 + bx + c & x \leq 1 \\ x + 1 & 1 \lt x \lt 3 \\ x^2 + bx + c & x \geq 3 \end{array} \right. }$$
Now we can see that we need to look at the two points $$x=1$$ and $$x=3$$.
To have continuity at $$x=1$$ we need the following limit to hold.
$$\begin{array}{rcl} \displaystyle{ \lim_{x \to 1^-}{ x^2+bx+c} } & = & \displaystyle{ \lim_{x \to 1^+}{ x+1 } } \\ 1^2 + b(1) + c & = & 1 + 1 \\ b+c & = & 1 \end{array}$$
This gives us one equation involving b and c. The other equation comes from this limit. $$\begin{array}{rcl} \displaystyle{ \lim_{x \to 3^-}{ x+1 } } & = & \displaystyle{ \lim_{x \to 3^+}{ x^2+bx+c } } \\ 3 + 1 & = & 9 + 3b + c \\ 3b + c & = & -5 \end{array}$$
Solving the two equations $$b+c = 1$$ and $$3b + c = -5$$, we get $$b = -3, c = 4$$.

$$b = -3, c = 4$$

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[10 points] For $$\displaystyle{ g(x) = \frac{3x^2+3x}{x^2-3x-4} }$$.
(a) Calculate the zeros of $$g(x)$$.
(b) Calculate all values of x where vertical asymptotes occur.
(c) List all x-values where $$g(x)$$ is discontinuous and identify the discontinuities as removable or non-removable.

Problem Statement

[10 points] For $$\displaystyle{ g(x) = \frac{3x^2+3x}{x^2-3x-4} }$$.
(a) Calculate the zeros of $$g(x)$$.
(b) Calculate all values of x where vertical asymptotes occur.
(c) List all x-values where $$g(x)$$ is discontinuous and identify the discontinuities as removable or non-removable.

Solution

(a) The zeroes of rational functions occur where the denominator is not zero and the numerator is zero. So we need to set the numerator equal zero, solve for the x-values and make sure that those values do not also make the denominator zero.

 $$3x^2+3x=0 \to 3x(x+1)=0$$ $$x=0$$ and $$x=-1$$ Check the denominator. $$x=0 \to$$ $$x^2-3x-4 \to -4$$ ✔ $$x=-1 \to$$ $$x^2-3x-4 \to 0$$ ✘
 Final Answer - Part (a) $$x=0$$ is the only zero of $$g(x)$$.

(b) Vertical asymptotes of rational functions occur where the denominator is zero and the numerator is not zero. So we need to set the denominator to zero, solve for the x-values and check the numerator at those points. We already know where the numerator is zero, so we can just compare values.

 $$x^2-3x-4=0$$ $$(x-4)(x+1)=0$$ $$x=4$$ and $$x=-1$$ Compare these values to what we found for the numerator. For $$x=-1$$, we found that this value also made the numerator zero. So it is not an asymptote. For $$x=4$$, this is not one of the values we found in part a that made the numerator zero. So it is an asymptote.
 Final Answer - Part (b) $$x=4$$ is the only vertical asymptote of $$g(x)$$.

(c) Discontinuities occur when the denominator is zero. It doesn't matter what is going on with the numerator. However, the values in the numerator tell us if the discontinuity is removable or non-removable.
For this problem, the discontinuities occur at $$x=-1$$ and $$x=4$$. We found these values in part b.
$$x=4$$ is a vertical asymptote, so it is non-removable.
Since the numerator and denominator are both zero at $$x=-1$$, this discontinuity is a hole and is therefore removable.

 Final Answer - Part (c) $$g(x)$$ has two discontinuities, $$x=-1$$ and $$x=4$$ $$x=-1$$ is a removable discontinuity $$x=4$$ is a non-removable discontinuity

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[15 points] Given the function $$\displaystyle{ g(x) = \left\{ \begin{array}{cl} 2x + 1 & x \lt -1 \\ -x^2 + 3x - 2.5 & x \geq -1 \end{array} \right. }$$
(a) Accurately sketch $$g(x)$$.
(b) Evaluate $$\displaystyle{ \lim_{x \to -1^-}{ g(x) } }$$
(c) Evaluate $$\displaystyle{ \lim_{x \to -1^+}{ g(x) } }$$
(d) Is $$g(x)$$ continuous at $$x=-1$$? Why or why not?

Problem Statement

[15 points] Given the function $$\displaystyle{ g(x) = \left\{ \begin{array}{cl} 2x + 1 & x \lt -1 \\ -x^2 + 3x - 2.5 & x \geq -1 \end{array} \right. }$$
(a) Accurately sketch $$g(x)$$.
(b) Evaluate $$\displaystyle{ \lim_{x \to -1^-}{ g(x) } }$$
(c) Evaluate $$\displaystyle{ \lim_{x \to -1^+}{ g(x) } }$$
(d) Is $$g(x)$$ continuous at $$x=-1$$? Why or why not?

Solution (a) The plot is shown here.
(b) $$\displaystyle{ \lim_{x \to -1^-}{ g(x) } }$$
Since we are looking on the left of $$x=-1$$, we need the equation associated with $$x \lt -1$$. So our limit is $$\displaystyle{ \lim_{x \to -1^-}{ 2x+1 } = 2(-1)+1 = -1 }$$

 Final Answer - Part (b) $$\displaystyle{ \lim_{x \to -1^-}{ g(x) } = -1 }$$

(c) $$\displaystyle{ \lim_{x \to -1^+}{ g(x) } }$$
Since we are looking on the right of $$x=-1$$, we need the equation associated with $$x \geq -1$$. So our limit is $$\displaystyle{ \lim_{x \to -1^+}{ -x^2 + 3x - 2.5 } = -(-1)^2 + 3(-1)-2.5 = -6.5 }$$

 Final Answer - Part (c) $$\displaystyle{ \lim_{x \to -1^+}{ g(x) } = -6.5 }$$

(d) No, the function is not continuous at $$x=-1$$ because the limits are different from the right and from the left, i.e. $$\displaystyle{ \lim_{x \to -1^-}{ g(x) } \neq \lim_{x \to -1^+}{ g(x) } }$$

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[15 points] For the function $$\displaystyle{ f(x) = \frac{1+x-2x^2-x^3}{x^2+1} }$$
(a) Evaluate $$\displaystyle{ \lim_{x \to -\infty}{ f(x) } }$$
(b) Evaluate $$\displaystyle{ \lim_{x \to \infty}{ f(x) } }$$
(c) Determine the end behaviour of $$f(x)$$.

Problem Statement

[15 points] For the function $$\displaystyle{ f(x) = \frac{1+x-2x^2-x^3}{x^2+1} }$$
(a) Evaluate $$\displaystyle{ \lim_{x \to -\infty}{ f(x) } }$$
(b) Evaluate $$\displaystyle{ \lim_{x \to \infty}{ f(x) } }$$
(c) Determine the end behaviour of $$f(x)$$.

Solution

 (a) $$\displaystyle{ \lim_{x \to -\infty}{ f(x) } }$$ $$\displaystyle{ \lim_{x \to -\infty}{ \left[ \frac{1+x-2x^2-x^3}{x^2+1} \frac{1/x^2}{1/x^2} \right] } }$$ $$\displaystyle{ \lim_{x \to -\infty}{ \left[ \frac{1/x^2 + 1/x - 2 - x}{1 + 1/x^2} \right] } }$$ $$\displaystyle{ \frac{0 + 0 - 2 + \infty}{1 + 0} = +\infty }$$
 Final Answer - Part (a) $$\displaystyle{ \lim_{x \to -\infty}{ f(x) } = +\infty }$$
 (b) $$\displaystyle{ \lim_{x \to \infty}{ f(x) } }$$ $$\displaystyle{ \lim_{x \to \infty}{ \left[ \frac{1+x-2x^2-x^3}{x^2+1} \frac{1/x^2}{1/x^2} \right] } }$$ $$\displaystyle{ \lim_{x \to \infty}{ \left[ \frac{1/x^2 + 1/x - 2 - x}{1 + 1/x^2} \right] } }$$ $$\displaystyle{ \frac{0 + 0 - 2 - \infty}{1 + 0} = -\infty }$$
 Final Answer - Part (b) $$\displaystyle{ \lim_{x \to \infty}{ f(x) } = -\infty }$$

(c) To evaluate the end behaviour we need to do long division on the polynomials. The result is $f(x) = \frac{1+x-2x^2-x^3}{x^2+1} = -x - 2 + \frac{2x+3}{x^2+1}$ As x approaches $$\pm \infty$$, the fraction goes to zero, leaving $$-x-2$$. This is the end behaviour, the slant asymptote $$y=-(x+2)$$.

 Final Answer - Part (c) slant asymptote $$y=-(x+2)$$

Although we were not asked to plot the function, it helps to understand what is going on when we are able to see a plot. So below we have plotted the function $$f(x)$$ (in green) and the slant asymptote (in red). Built with GeoGebra

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[10 points] Evaluate $$\displaystyle{ \lim_{x \to 3}{ \frac{\sqrt{x+13}-4}{x-3} } }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \lim_{x \to 3}{ \frac{\sqrt{x+13}-4}{x-3} } }$$

$$\displaystyle{ \lim_{x \to 3}{ \frac{\sqrt{x+13}-4}{x-3} } = \frac{1}{8} }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \lim_{x \to 3}{ \frac{\sqrt{x+13}-4}{x-3} } }$$

Solution

Direct substitution yields $$0/0$$ which is indeterminate. So we will rationalize the fraction by multiplying the numerator and denominator by $$\sqrt{x+13}+4$$.

 $$\displaystyle{ \lim_{x \to 3}{ \left[ \frac{\sqrt{x+13}-4}{x-3} \frac{\sqrt{x+13}+4}{\sqrt{x+13}+4} \right] } }$$ $$\displaystyle{ \lim_{x \to 3}{ \frac{x-3}{(x-3)(\sqrt{x+13}+4)} } }$$ $$\displaystyle{ \lim_{x \to 3}{ \frac{1}{(\sqrt{x+13}+4)} } }$$ $$\displaystyle{ \frac{1}{\sqrt{16}+4} }$$ $$\displaystyle{ \frac{1}{8} }$$

$$\displaystyle{ \lim_{x \to 3}{ \frac{\sqrt{x+13}-4}{x-3} } = \frac{1}{8} }$$

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[10 points] Let $$s(t) = -5t^2 + 175$$ be the position function, which gives the height in meters of an object. The velocity at time $$t=a$$ seconds is given by $\lim_{t \to a}{ \frac{s(a)-s(t)}{a-t} }$ (a) Using the limit, find the general equation for the velocity as a function of t.
(b) Using your answer from part (a), calculate the velocity of the object when $$t=1.5$$ seconds.
(c) Using your answer from part (a), calculate the velocity of the object when it hits the ground.

Problem Statement

[10 points] Let $$s(t) = -5t^2 + 175$$ be the position function, which gives the height in meters of an object. The velocity at time $$t=a$$ seconds is given by $\lim_{t \to a}{ \frac{s(a)-s(t)}{a-t} }$ (a) Using the limit, find the general equation for the velocity as a function of t.
(b) Using your answer from part (a), calculate the velocity of the object when $$t=1.5$$ seconds.
(c) Using your answer from part (a), calculate the velocity of the object when it hits the ground.

Solution

(a) calculate $$v(t)$$ from the limit

 $$\displaystyle{ \lim_{t \to a}{ \frac{s(a)-s(t)}{a-t} } }$$ $$s(a) = -5a^2 + 175$$ $$\displaystyle{ \lim_{t \to a}{ \frac{(-5a^2 + 175 )-(-5t^2 + 175 )}{a-t} } }$$ $$\displaystyle{ \lim_{t \to a}{ \frac{-5a^2 + 175 +5t^2 - 175 }{a-t} } }$$ $$\displaystyle{ \lim_{t \to a}{ \frac{-5(a^2 - t^2)}{a-t} } }$$ $$\displaystyle{ \lim_{t \to a}{ \frac{-5(a-t)(a+t)}{a-t} } }$$ $$\displaystyle{ \lim_{t \to a}{ -5(a+t) } }$$ $$-10a$$
 Final Answer - Part (a) $$v(t) = -10t$$ m/sec

(b) calculate velocity at t=1.5sec
$$v(1.5) = -10(1.5) = -15$$ m/sec

 Final Answer - Part (b) $$v(1.5) = -15$$ m/sec

(c) calculate velocity when object hits the ground
The oject hits the ground when the height is zero. So we need to solve the position function for the time when height is zero.
$$\begin{array}{rcl} -5t^2 + 175 & = & 0 \\ - 5t^2 & = & -175 \\ t^2 & = & 35 \\ t & = & \pm \sqrt{35} \end{array}$$
Since we assume time is positive, we choose the time $$t=\sqrt{35}$$. Now we need to put that value in the velocity equation to get our answer.

 Final Answer - Part (c) $$v(\sqrt{35}) = -10\sqrt{35}$$ m/sec

Note: Velocity is negative since the object is moving down, in the negative direction.

Log in to rate this practice problem and to see it's current rating. Plot of $$f(x)$$

[40 points] Use the plot of $$f(x)$$ to answer these questions. You do not need to explain your answers.
(a) For each of the a-values $$\{ -4, -2, -1, 1, 5, 6 \}$$ evaluate $$\displaystyle{ \lim_{x \to a^-}{ f(x) } }$$, $$\displaystyle{ \lim_{x \to a^+}{ f(x) } }$$, $$\displaystyle{ \lim_{x \to a}{ f(x) } }$$, $$f(a)$$ and answer the question: Is $$f(x)$$ continuous at $$x=a$$?
(b) List all x-values where $$f(x)$$ has discontinuites and state whether each is removable or non-removable.
(c) List all vertical and horizontal asymptotes, specifying which type of asymptote it is.
(d) State the domain and range of $$f(x)$$.

Problem Statement Plot of $$f(x)$$

[40 points] Use the plot of $$f(x)$$ to answer these questions. You do not need to explain your answers.
(a) For each of the a-values $$\{ -4, -2, -1, 1, 5, 6 \}$$ evaluate $$\displaystyle{ \lim_{x \to a^-}{ f(x) } }$$, $$\displaystyle{ \lim_{x \to a^+}{ f(x) } }$$, $$\displaystyle{ \lim_{x \to a}{ f(x) } }$$, $$f(a)$$ and answer the question: Is $$f(x)$$ continuous at $$x=a$$?
(b) List all x-values where $$f(x)$$ has discontinuites and state whether each is removable or non-removable.
(c) List all vertical and horizontal asymptotes, specifying which type of asymptote it is.
(d) State the domain and range of $$f(x)$$.

Solution

(a) limits and continuity

 $$\displaystyle{ \lim_{x \to -4^-}{ f(x) } = 0 }$$ $$\displaystyle{ \lim_{x \to -4^+}{ f(x) } = +\infty }$$ $$\displaystyle{ \lim_{x \to -4}{ f(x) } = }$$ DNE $$f(-4) =$$ DNE Is $$f(x)$$ continuous at $$x=-4$$? No
 $$\displaystyle{ \lim_{x \to -2^-}{ f(x) } = 2 }$$ $$\displaystyle{ \lim_{x \to -2^+}{ f(x) } = 5 }$$ $$\displaystyle{ \lim_{x \to -2}{ f(x) } = }$$ DNE $$f(-2) = 1$$ Is $$f(x)$$ continuous at $$x=-2$$? No
 $$\displaystyle{ \lim_{x \to -1^-}{ f(x) } = 2 }$$ $$\displaystyle{ \lim_{x \to -1^+}{ f(x) } = 2 }$$ $$\displaystyle{ \lim_{x \to -1}{ f(x) } = 2 }$$ $$f(-1) = 2$$ Is $$f(x)$$ continuous at $$x=-1$$? Yes
 $$\displaystyle{ \lim_{x \to 1^-}{ f(x) } = 2 }$$ $$\displaystyle{ \lim_{x \to 1^+}{ f(x) } = -1 }$$ $$\displaystyle{ \lim_{x \to 1}{ f(x) } = }$$ DNE $$f(1) = -1$$ Is $$f(x)$$ continuous at $$x=1$$? No
 $$\displaystyle{ \lim_{x \to 5^-}{ f(x) } = -1 }$$ $$\displaystyle{ \lim_{x \to 5^+}{ f(x) } = -1 }$$ $$\displaystyle{ \lim_{x \to 5}{ f(x) } = -1 }$$ $$f(5) =$$ DNE Is $$f(x)$$ continuous at $$x=5$$? No
 $$\displaystyle{ \lim_{x \to 6^-}{ f(x) } = -\infty }$$ $$\displaystyle{ \lim_{x \to 6^+}{ f(x) } = -\infty }$$ $$\displaystyle{ \lim_{x \to 6}{ f(x) } = -\infty }$$ $$f(6) = -1$$ Is $$f(x)$$ continuous at $$x=6$$? No

(b) discontinuities

$$x=-4$$

non-removable

$$x=-2$$

non-removable

$$x=1$$

non-removable

$$x=5$$

removable

$$x=6$$

non-removable

(c) asymptotes

$$x=-4$$

vertical

$$x=6$$

vertical

$$y=0$$

horizontal

(d) domain and range

all reals except for $$x=-4$$

domain

all reals except for $$0 \leq y \lt 1$$

range

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### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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