Respuesta :
Answer:
Step-by-step explanation:
the function you have given is not well stated out since it is not division or multiplication. however i will use the function
[tex]f(x) = \frac{x}{x+6}[/tex] and i believe with this example you will get a better view on how to answer similar questions.
for f(x) to satisfy the Mean Value Theorem, it must be continuous in the closed interval [1, 12] and differentiable in the open interval (1, 12).
f(x) is undefined if x= -6 ∉[1, 12]. hence f(x) is continuous on the closed interval [1, 12].
next we check if f(x) is differentiable in the open interval (1 , 12).
using quotient rule, we have
[tex]\frac{d}{dx} \frac{h(x)}{g(x)} = \frac{h'(x)g(x) - h(x)g'(x)}{g^{2}(x) }[/tex]
= [tex]\frac{(1)(x+6)- (x)(1)}{(x+6)^{2} } = \frac{6}{(x+6)^2}[/tex]
at x=-6, f'(x) is undefined but -6∉( 1 , 12). thus f(x) is differentiable. hence the function [tex]f(x) = \frac{x}{x + 6}[/tex] satisfies Mean Value Theorem
- This shows that the function satisfies the Mean value theorem in the closed interval [1, 12] except at x = 6
- This shows that the function is differentiable in the closed interval (1, 12) except at x = 6
From the given question, we are to determine if the function f(x) is [tex]\frac{x}{x+6}[/tex] satisfies the mean value theorem
For f(x) to satisfy the Mean Value Theorem, it must be continuous in the closed interval [1, 12] and differentiable in the open interval (1, 12).
The function is not continuous when the denominator is equal to zero i.e.
[tex]x+6=0\\x=-6[/tex]
This shows that the function satisfies the Mean value theorem in the closed interval [1, 12] except at x = 6
For the function to be differentiable in the interval, we will need to differentiate the function using the quotient rule to get f'(x)
[tex]f'(x)=\frac{(x+6)(1)-x(1)}{(x+6)^2}\\f'(x)=\frac{x+6-x}{(x+6)^2}\\f'(x)= \frac{6}{(x+6)^2}[/tex]
This shows that the function is differentiable in the closed interval (1, 12) except at x = 6
Learn more here: https://brainly.com/question/12529002
