Answer:
Any value of c.
Step-by-step explanation:
The MVT applied on this interval states the following:
There exists some c∈(0,4) such that [tex]f'(c)=\frac{f(4)-f(0)}{4-0}=frac{f(4)-4}{4}=\frac{3(4)+4-4}{4}=3[/tex].
Thus, we must solve the equation [tex]f'(c)=3[/tex]. Using derivative rules, [tex]f'(c)=(3c+4)'=3[/tex]. Then the equation becomes [tex]3=3[/tex] which is true for all c∈(0,4).
Another way of interpreting this is the following: the graph of f is a line with slope equal to 3, which is equal to [tex]\frac{f(4)-f(0)}{4-0}[/tex] (by the slope formula). But f'(c) is the slope of the tangent line to f at c, and the tangent line is f itself. Thus f'(c)=3 for all c.
