4.4 The mean value theorem  (Page 2/7)

The mean value theorem and its meaning

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $[a,b]$ and differentiable over the open interval $\left(a,b\right),$ then there exists a point $c\in \left(a,b\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$

Proof

The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$ Since the slope of that line is

and the line passes through the point $\left(a,f\left(a\right)\right),$ the equation of that line can be written as

Let $g\left(x\right)$ denote the vertical difference between the point $\left(x,f\left(x\right)\right)$ and the point $\left(x,y\right)$ on that line. Therefore,

Since the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\left(a\right)=0=g\left(b\right).$ Since $f$ is a differentiable function over $\left(a,b\right),$ $g$ is also a differentiable function over $\left(a,b\right).$ Furthermore, since $f$ is continuous over $[a,b],$ $g$ is also continuous over $[a,b].$ Therefore, $g$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $c\in \left(a,b\right)$ such that $g\prime \left(c\right)=0.$ Since

we see that

Since $g\prime (c)=0,$ we conclude that

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In the next example, we show how the Mean Value Theorem can be applied to the function $f(x)=\sqrt{x}$ over the interval $[0,9].$ The method is the same for other functions, although sometimes with more interesting consequences.

Verifying that the mean value theorem applies

For $f\left(x\right)=\sqrt{x}$ over the interval $[0,9],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\in (0,9)$ such that $f^{\prime }(c)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.

We know that $f(x)=\sqrt{x}$ is continuous over $[0,9]$ and differentiable over $\left(0,9\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\in \left(0,9\right)$ such that $f^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative $f^{\prime }\left(x\right)=\frac{1}{(2\sqrt{x})}.$ The slope of the line connecting $(0,f(0))$ and $(9,f(9))$ is given by

$\frac{f(9)-f(0)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.$

We want to find $c$ such that $f^{\prime }(c)=\frac{1}{3}.$ That is, we want to find $c$ such that

$\frac{1}{2\sqrt{c}}=\frac{1}{3}.$

Solving this equation for $c,$ we obtain $c=\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.

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