Respuesta :
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
[tex]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} [/tex]
4∈[2,8] is the required point.
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
[tex]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} [/tex]
4∈[2,8] is the required point.
Answer:
[tex]c=4[/tex]
Step-by-step explanation:
By defintion, the mean value theorem is
[tex]f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]
So, in this case, we know that [tex]a=2[/tex] and [tex]b=8[/tex].
Now, we need to find [tex]f(8)[/tex] and [tex]f(2)[/tex], by replacing those values into the given function
[tex]f(x)=10-\frac{16}{x}\\ f(8)=10-\frac{16}{8}=10-2=8[/tex]
So, [tex]f(b)=f(8)=8[/tex].
[tex]f(x)=10-\frac{16}{x}\\ f(2)=10-\frac{16}{2}=10-8=2[/tex]
So, [tex]f(2)=2[/tex]
Then, we replace all values,
[tex]f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{8-2}{8-2}=\frac{6}{6}=1[/tex]
Now, we find the derivative of the function which has to be equal to one,
[tex]f(x)=10-\frac{16}{x}=10-16x^{-1} \\f'(x)=16x^{-2} =\frac{16}{x^{2} } =1[/tex]
Now, we solve the equation to find the value c,
[tex]\frac{16}{x^{2} } =1\\x^{2} =16\\x=4[/tex]
Therefore, [tex]c=4[/tex] is the value inside the given interval that satisfy the mean value theorem.
