The Mean Value Theorem
MEAN VALUE THEOREM
Conditions:
2. The function f must be differentiable on the open interval (a, b).
If these two conditions are met then there is a number c in the open interval (a, b) such that:
${f}^{\prime}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
Let's use this theorem in a couple examples.
$f\left(x\right)={x}^{3}+4,\left[-1,1\right]$
Step 1: Determine if the Mean Value Theorem can be applied. |
To apply the Mean Value Theorem the function must be continuous on the closed interval and differentiable on the open interval. This function is a polynomial function, which is both continuous and differentiable on the entire real number line and thus meets these conditions. |
Step 2: Find the slope of the secant line. |
Determine f(a) if a = -1, the left endpoint of the interval. $f\left(-1\right)={\left(-1\right)}^{3}+4=3$ $\left(a,f\left(a\right)\right)=\left(-1,3\right)$ Determine f(b) if b = 1, the right endpoint of the interval. $f\left(1\right)={\left(1\right)}^{3}+4=5$ $\left(b,f\left(b\right)\right)=\left(1,5\right)$ Determine the slope of the secant line between -1 and 1. ${m}=\frac{\Delta y}{\Delta x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{5-3}{1-\left(-1\right)}=\frac{2}{2}={1}$ |
Step 3: Determine the first derivative of the function. |
f(x) = x^{3} + 4 ${f}^{\prime}\left(x\right)=3{x}^{2}$ |
Step 4: Set ${f}^{\prime}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ and solve for c. ${f}^{\prime}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ Mean Value Theorem ${f}^{\prime}\left(c\right)=1$ Sub. from Step 2 3c^{2} = 1 Sub. for $f\prime \left(c\right)$ ${c}^{2}=\frac{1}{3}$ Divide by 3 $c=\pm \frac{1}{\sqrt{3}}$ Square root $f\prime \left(\frac{1}{\sqrt{3}}\right)=1$ and $f\prime \left(-\frac{1}{\sqrt{3}}\right)=1$
There are two values of c in the interval of [-1, 1] for which the tangent line of c is parallel to the secant line between -1 and 1. |
$f\left(x\right)=\frac{x+3}{x},\left[1,3\right]$
Step 1: Determine if the Mean Value Theorem can be applied. |
To apply the Mean Value Theorem the function must be continuous on the closed interval and differentiable on the open interval. This function is a rational function, which is both continuous on the interval [1, 3]and differentiable on the interval (1, 3). |
Step 2: Find the slope of the secant line. |
Determine f(a) if a = 1, the left endpoint of the interval. $f\left(1\right)=\frac{1+3}{1}=4$ $\left(a,f\left(a\right)\right)=\left(1,4\right)$ Determine f(b) if b = 3, the right endpoint of the interval. $f\left(3\right)=\frac{3+3}{3}=2$ $\left(b,f\left(b\right)\right)=\left(3,2\right)$ Determine the slope of the secant line. ${m}=\frac{\Delta y}{\Delta x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{2-4}{3-1}=\frac{-2}{2}={-}{1}$ |
Step 3: Determine the first derivative of the function. |
$f\left(x\right)=\left(x+3\right)\left({x}^{-1}\right)=1+3{x}^{-1}$ ${f}^{\prime}\left(x\right)={-}{3}{x}^{-2}$ |
Step 4: Set ${f}^{\prime}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ and solve for c. ${f}^{\prime}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ Mean Value Theorem ${f}^{\prime}\left(c\right)=-1$ Sub. from Step 2 $-3{c}^{-2}=-1$ Sub. for $f\prime \left(c\right)$ ${c}^{-2}=\frac{1}{3}$ Divide by -3 c^{2} = 3 Take reciprocal $c=\pm \sqrt{3}$ Square root $f\prime \left(\sqrt{3}\right)=-1$ and $f\prime \left(-\sqrt{3}\right)=1$ (but this point is not in the interval)
There is only one value for c in the interval of [1, 3] for which the tangent line of c is parallel to the secant line between 1 and 3. |
Related Links: Math algebra Limits: Introduction and One-Sided Limits Limits: Infinite Limits Calculus Topics |
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