17Calculus - Intermediate Value Theorem

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The intermediate value theorem is used to establish that a function passes through a certain y-value and relies heavily on continuity. First, let's look at the theorem itself.

Intermediate Value Theorem

For a continuous function, $$f(x)$$ on an interval $$[a,b]$$, if $$t$$ is between $$f(a)$$ and $$f(b)$$,
then there exists a value $$c \in [a,b]$$ such that $$f(c) = t$$.

Notice that the theorem just tells you that the value $$x=c$$ exists but doesn't tell you what it is or how to find it.
To get the idea of this theorem clear in your head, here are some great videos for you to watch. They use graphs to help you understand what the theorem means.

Here is a video that shows, graphically, how the intermediate value theorem works. She uses color in her graph to make it easy to follow.

Krista King Math - Intermediate Value Theorem [4min-5secs]

video by Krista King Math

Here is a great video that clearly explains the intermediate value theorem more from a mathematical point of view than in the previous video.

PatrickJMT - Intermediate Value Theorem [7min-53secs]

video by PatrickJMT

Application of the Intermediate Value Theorem - - Here is a great video showing a non-standard application of the IVT. To work this problem, he uses the definition of the limit. Don't skip this video. It will help you understand limits, continuity and the IVT.

Dr Chris Tisdell - IVT [5min-58secs]

video by Dr Chris Tisdell

Practice

Use the intermediate value theorem to solve these problems. Give your answers in exact form.

Show that $$x^3-3x+1=0$$ has a root in the interval $$(0,1)$$.

Problem Statement

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that $$x^3-3x+1=0$$ has a root in the interval $$(0,1)$$.

Solution

415 video

video by PatrickJMT

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Show that $$f(x)=x^4+x-3$$ has a root in the interval $$(1,2)$$.

Problem Statement

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that $$f(x)=x^4+x-3$$ has a root in the interval $$(1,2)$$.

Solution

422 video

video by Krista King Math

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Show that $$x^2=\sqrt{x+1}$$ has a root in the interval $$(1,2)$$.

Problem Statement

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that $$x^2=\sqrt{x+1}$$ has a root in the interval $$(1,2)$$.

Solution

416 video

video by PatrickJMT

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Show that $$e^x=2\cos(x)$$ has at least one positive root.

Problem Statement

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that $$e^x=2\cos(x)$$ has at least one positive root.

Solution

He chose $$x=0$$ and $$x=\pi/2$$ to show that the function has a positive value and negative value. You can choose other values as long as one results in a positive value and the other is negative.

414 video

video by Dr Chris Tisdell

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Prove that $$\cos(x)=x^3$$ has at least one real root.

Problem Statement

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Prove that $$\cos(x)=x^3$$ has at least one real root.

Solution

421 video

video by Krista King Math

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intermediate value theorem 17calculus youtube playlist

You CAN Ace Calculus

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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

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

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 Intermediate Value Theorem Practice

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