Answer:
1)[tex]a<c<b\\f'(c)=\frac{f(4)-f(3)}{4-3}\Rightarrow f'(c)=f(4)-f(3)[/tex]
2)[tex]c=f'(c)\\f'(c)=\int_{4}^{3}f'(x)dx[/tex]
Step-by-step explanation:
Despite some missing part like the function, still we can affirm:
1) The Mean Value Theorem states that given a differentiable function f over an interval (a, b) [in this case (3,4)], then we know that there's a value "c" of between the endpoints of this interval.
And this c:
[tex]a<c<b\\f'(c)=\frac{f(4)-f(3)}{4-3}\Rightarrow f'(c)=f(4)-f(3)[/tex]
2) If c equals to the average value, then we can say that:
[tex]c=f'(c)[/tex]
And
[tex]f'(c)=\frac{1}{4-3}\int_{4}^{3}\frac{f'(x)}{4-3}dx\Rightarrow f'(c)=\int_{4}^{3}f'(x)dx[/tex]
