Given the curve $y = 2 x ^3 + 3 x ^2 − 36 x$ on the interval $− 1 \le x \le4$ , find the absolute minimum?

Question

I know I have to find the first derivative which is $y'=6x^2+6x-36.$ Then I set it to $0.$ Which I got $0 = 6(x^2+x-6)$ and then $x=-3$ or $x=2.$

Answer 1

Hint:

You also know that this derivative is $>0$ if $|x|$ is large enough, so, on $\mathbf R$, the function is increasing for $x<-3$, decreasing if $-3<x<2$, and finally increasing if $x>2$. Therefore, on the interval $[-1,4]$, the function has an absolute minimum at $x=2$.

Answer 2

This isn't a complete answer, but is guidance towards a complete answer.

You're doing great. When you want to find the max/min on a closed interval like these, there are only a few kinds of $x$-values to consider:

1. Those where $f'(x) = 0$, which you've found.

2. Those where $f'(x)$ doesn't exist (think of a graph that looks like the letter "V" -- the min happens at a point where there's no derivative!)

3. The end points.

So you've found those in the first category -- there's just $x = 2$. There are none in the second group. In the third group, there's $-1$ and $4$.

At each of these $x$-values, you need to look at $f(x)$. For instance, we have $f(-1) = -2 + 3 + 36 = 38$. How does that compare to $f(2)$ and $f(4)$?

Answer 3

Absolute minimum occurs either at the boundary points or at the points where the derivative equals to zero or does not exist. So, you've already found $x=2$ is a point in the interval such that the derivative is $0$. You would only need to check the function's value at $x=2$ and the boundary points $x=-1,x=4$ to determine the absolute minimum point in that interval.