Find the $x$ at which the local maxima of a function ocours.

Question

The function $f(x) = \int\limits_{-1}^{x}t(e^t-1)(t-1)(t-2)^3(t-3)^5 dt$ has a local maxima at $x=?$

First, I differentiated $f(x)$ and found its roots.

That came out to be $x = 0,1,2,3$. Now, one of those numbers, when plugged into $f(x)$, must give the largest value compared to the others.

In our case, since we are integrating, we would need to find the $x$ that minimizes the negative area of the function that we are integrating.

By further work and analyzing the function, I figured out that the answer must be $0$ or $2.$

But, I don't know what to do next?

Any help would be appreciated.

Answer 1

HINT:

The sign of $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5$$ is $-,+,-,+$ respectively on the intervals $(-1,1),(1,2),(2,3),(3,\infty).$ From this you can finish without further calculation.

Answer 2

You differentiated and found the roots correctly: $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5=0 \Rightarrow \\ x_0=\{0,1,2,3\};$$ Method 1. Check the neighborhood of the critical points: $$\begin{array}{c|c|c|c} x_0&f'(x_0-\epsilon)&f'(x_0+\epsilon)&\text{Type of stationary point}\\ \hline 0&-&-&\text{inflection}\\ 1&-&+&\text{local minimum}\\ 2&+&-&\text{local maximum}\\ 3&-&+&\text{local minimum}\\ \end{array}$$

Method 2. Check the second (or higher) order derivative at the critical points: $$\begin{align}f''(0)&=0 \ \ \ \text{(inconclusive)}\\ f''(1)&=32(e-1)>0 \ \ \text{(local minimum)}\\ f''(2)&=0 \ \ \text{(inconclusive)}\\ f''(3)&=0 \ \ \text{(inconclusive)}\\ \end{align}$$ Now you can use higher order derivative test: $$\begin{align}f'''(0)&<0 \ \ \text{(strictly decreasing inflection point)}\\ f'''(2)&=0, f^{(4)}(2)<0 \ \ \text{(local maximum)}\\ f'''(3)&=f^{(4)}(3)=f^{(5)}(3)=0, f^{(6)}(3)>0 \ \ \text{(local minimum)}.\end{align}$$ Note: You don't have to calculate the higher order derivatives completely, but you should focus on the corresponding factor and the resulting sign.