Verified Answer :
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:
\(c=4\)
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.