How to Work with the Intermediate Value Theorem?

A step-by-step guide to working with the intermediate value theorem

Working with the Intermediate Value Theorem – Example 1:

First, find the values of the given function at the \(x=0\) and \(x=2\).

Substitute \(x=0\):

\(f(x)=x^5-2x^3-2=0\) → \(f(0)=(0)^5-2(0)^3-2\)

\(f(0)=-2\)

Substitute \(x=2\):

\(f(x)=x^5-2x^3-2=0\) → \(f(2)=(2)^5-2(2)^3-2\)

\(f(2)=32-16-2\)

\(f(2)=14\)

Therefore, we conclude that at \(x = 0\), the curve is below zero; while at \(x = 2\), it is above zero.

Since the given equation is polynomial, its graph will be continuous. Therefore, by applying the intermediate value theorem, we can say that the graph should cross at some point between \([0, 2]\).

Exercises for Working with the Intermediate Value Theorem

1. The function \(h(x)\) is continuous on the interval \((1,8)\). If \(h(1)=-7\) and \(h(8)=-6\) can you conclude that \(h(x)\) is ever equal to \(0\)?
2. For \(f(x)=\frac{1}{x}\), \(f(-1)=-1< 0\) and \(f(1)=1>0\). Can we conclude that \(f(x)\) has a zero in the interval \([-1,1]\)?
3. Can we use the intermediate value theorem to conclude that \(f(x)=sin x\) equals \(0.4\) at some place in the interval \([\frac{\pi}{2},\pi]\)?

1. \(\color{blue}{No}\)
2. \(\color{blue}{No}\)
3. \(\color{blue}{Yes}\)

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