Answer to Question #310474 in Calculus for Mom

Question

Estimate the values of c that satisfy the conclusion of the Mean Value Theorem on the interval (0,8). Enter your answer as a comma-separated list. Round to one decimal.

Answer 1

Solution

According to the mean value theorem,

If  $f(x)$ is continuous on interval $[a, b]$ , and differentiable on $(a, b)$ .

Then there exists $c$ such that $a < c <b$ and 

$f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$

In the question the function is missing, so let us choose, the function as

$f\left( x \right) = {x^3} + 2x - 3$ which is continuous with in the interval $[0,8]$ and differentiable within $(0, 8)$ ,

then for $0 < c <8$, we have

$f'\left( c \right) = \frac{{f\left( 8 \right) - f\left( 0 \right)}}{{8 - 0}}$

$f'\left( c \right) = \frac{{\left[ {{{\left( 0 \right)}^3} - 2\left( 0 \right) - 3} \right] - \left[ {{{\left( 8 \right)}^3} - 2\left( 8 \right) - 3} \right]}}{{8 - 0}}\\$

$f'\left( c \right) = \frac{{ - 3 - 493}}{8}\\$

$f'\left( c \right) = 62$

Now

$f\left( x \right) = {x^3} + 2x - 3\\$

$f'\left( x \right) = 3{x^2} + 2\\$

$f'\left( c \right) = 3{c^2} + 2$

Therefore,

$3{c^2} + 2=62$

$3{c^2} =62-2$

$c^2=\frac{60}{3}=20$

$c=2\sqrt 5 \$ (ignoring negative value since we need to focus within the interval $[0, 8]$ )

since we can see that

$0<2\sqrt 5 < 8$

Hence required value of $c$ is $2\sqrt 5$