Solutions

a) For contradiction, suppose $g(a)$ and $g(b)$ have different signs. Since $f$ is a polynomial function and is therefore continuous, and since the sum of continuous functions, which a polynomial function is expressed as, is continuous, then $g$ is continuous. We can thus apply the Intermediate Value Theorem and assert that there exists an $x\in(a,b)$ such that $g(x)=0$. However, $x$ would be a root not only of $g$, but also of $f$. This contradicts the premise that $a$ and $b$ are consecutive roots.

b) We have that

$$f'(x)=(x-b)g(x)+(x-a)g(x)+(x-a)(x-b)g'(x).$$

Hence,

$$f'(a)=(a-b)g(a)$$

and

$$f'(b)=(b-a)g(b).$$

Since we've established in a) that the signs of $g(a)$ and $g(b)$ are the same, then the signs of $f'(a)$ and $f'(b)$ are different. Note also that $a\ne b$ because they're not double roots.

We can therefore reapply the Intermediate Value Theorem to assert that there exists $c\in(a,b)$ such that $f'(c)=0$.

c) We have that

$$f'(x)=m(x-a)^{m-1}(x-b)^ng(x)+(x-a)^mn(x-b)^{n-1}g(x)+(x-a)^m(x-b)^ng'(x).$$

Moreover,

$$h(x)=\frac{f'(x)}{(x-a)^{m-1}(x-b)^{n-1}}=m(x-b)g(x)+n(x-a)g(x)+(x-a)(x-b)g'(x).$$

Thus, $h(a)=m(a-b)g(a)$ and $h(b)=n(b-a)g(b)$. Since their signs are different, we can apply the Intermediate Value Theorem and assert that there exists $c\in(a,b)$ such that $h(c)=0$ and therefore $f'(c)=0$.