can someone please tell me if I'm going in the right direction?
Find $\max_{x\in[-3, 3]} \{bx^2 - 3x + b^2\} $ where $b$ is a real parameter.
Here is how I reasoned: $[-3, 3]$ is a closed and bounded set, hence it's compact. By Weierstrass Theorem, being $f(x)$ a polynomial hence continuous in the whole real line and particularly in $[-3, 3]$, the existence of an absolute max / min in $[-3, 3]$ is guaranteed.
So now $f'(x) = 2bx - 3 = 0$ and its zeroes are given by $x_c = \frac{3}{2b}$
Now I thought I have to find $b$ such that $x_c \in [-3, 3]$ and I found that I need $b\in [-1/2, 1/2]$
That being said, I wanted to compare $f$ in the extrema of the interval and $f(x_c)$:
$$f(-3) = 9b + 6 + b^2$$ $$f(3) = 9b - 6 + b^2$$ $$f(x_c) = \frac{4b^3-9}{4b}$$
Reasoning Update
I thought to compare the images of the three points. I mean I did:
$$f(3) > f(-3)$$
This turned out to be never true hence $f(-3) > f(3)$ for all $x$ and for all $b$
In the same way I did
$$f(x_c) < f(-3)$$
And I found out that this is true for all $x$ and $b$.
Hence I thought the function starts from $-3$ in a high point, decreasing down to $x_c$ which then is a minimum, and increasing up to $3$.
Hence $-3$ and $+3$ are the max, but the absolute is at $-3$.
Am I right?
Just one note: either typo $f(x) = bx^2 -3x + b$, or derivative is $f'(x) = 2bx = 0$.