Question

Given f(x) = 10 - 16/x, find all c in the interval [2,8] that satisfy the mean value theorem

Answer 1

F(x)=10-16/x
f'(x)=16/x^2
f'(c)=16/c^2
f(8)=10-16/8=10-2=8
f(2)=10-16/2=10-8=2
$f'(c)= \frac{f(b)-f(a)}{b-a} in (2,8), \frac{16}{c^2} = \frac{f(8)-f(2)}{8-2} , \frac{16}{c^2} = \frac{8-2}{8-2}$
4∈[2,8] is the required point.

Answer 2

Answer:

$c=4$

Step-by-step explanation:

By defintion, the mean value theorem is

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

So, in this case, we know that $a=2$ and $b=8$.

Now, we need to find $f(8)$ and $f(2)$, by replacing those values into the given function

$f(x)=10-\frac{16}{x}\\ f(8)=10-\frac{16}{8}=10-2=8$

So, $f(b)=f(8)=8$.

$f(x)=10-\frac{16}{x}\\ f(2)=10-\frac{16}{2}=10-8=2$

So, $f(2)=2$

Then, we replace all values,

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{8-2}{8-2}=\frac{6}{6}=1$

Now, we find the derivative of the function which has to be equal to one,

$f(x)=10-\frac{16}{x}=10-16x^{-1} \\f'(x)=16x^{-2} =\frac{16}{x^{2} } =1$

Now, we solve the equation to find the value c,

$\frac{16}{x^{2} } =1\\x^{2} =16\\x=4$

Therefore, $c=4$ is the value inside the given interval that satisfy the mean value theorem.