Question

Verify Rolle's theorem for the function f(x) = x² + 5x + 6 on the interval [−3, −2].

Answer 1

Given function is $f\left(x\right)=x^2+5x+6$.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on $\left[-3,-2\right]$.
Also,
$$\begin{gathered} f\left(-3\right)={\left(-3\right)}^2+5\left(-3\right)+6=9-15+6=0 \\ f\left(-2\right)={\left(-2\right)}^2+5\left(-2\right)+6=4-10+6=0 \\ \therefore f\left(-3\right)=f\left(-2\right)=0 \end{gathered}$$

Thus, all the conditions of the Rolle's theorem are satisfied.

Now, we have to show that there exists $c\in \left[-3,-2\right]$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=x^2+5x+6 \\ \Rightarrow f\text{'}\left(x\right)=2x+5 \\ \therefore f\text{'}\left(x\right)=0\Rightarrow 2x+5=0 \\ \Rightarrow x=\frac{-5}{2} \end{gathered}$$

Thus, $c=\frac{-5}{2}\in \left(-3,-2\right)\mathrm{such}\mathrm{that}f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.