Question

for g(x)=(x-4)/(x-3), we can use the mean value theorem on [4,6]. hence determine c

Answer 1

According to the mean value theorem: Suppose f(x) is a function that satisfies both of the following.

1. f(x) is continuous on the closed interval [a;b]
2. f(x) is differentiable on the open interval (a;b).

Then there is a number c such that a < c < b and

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

Function g(x)=(x-4)/(x-3) is continuous on the closed interval [4,6] and is differentiable on the open interval (4;6).

Therefore

$$g'(c)=\frac {g(6)-g(4)} {6-4}.$$

$$g(6)= \frac {6-4} {6-3}= \frac {2} {3}$$

$$g(4)= \frac {4-4} {4-3}= 0$$

As result get

$$g'(c)=\frac {\frac {2} {3} -0} {6-4}=\frac {1} {3}$$

On the other hand

$$g'(x)=(\frac {x-4} {x-3})'=\frac {1*(x-3)-1*(x-4)} {(x-3)^2}$$

$$g'(x)=\frac {x-3-x+4} {(x-3)^2}=\frac {1} {(x-3)^2}$$

Therefore

$$g'(c)=\frac {1} {(c-3)^2}=\frac {1} {3}$$

Solve the equation

$$(с-3)^2=3$$

$$c-3=\sqrt {3} \Longrightarrow с=3+\sqrt {3} \isin [4;6]$$

$$c-3=-\sqrt {3} \Longrightarrow с=3-\sqrt {3} \notin [4;6]$$

Hence

$$с=3+\sqrt {3}$$

Answer.

$$с=3+\sqrt {3}$$