Find the greatest value of the function $f(x) = x^4 - 6bx^2 + b^2$

Question

Find the greatest value of the function $f(x) = x^4 - 6bx^2 + b^2$ on the interval $[-2, 1]$ depending on the parameter $b$.

Answer 1

We note that on closed intervals, a continuous function (which polynomials are) must take extreme value at endpoints, or where the derivative is 0.

So, we solve

$$f'(x) = 4x^3 - 12bx = 0$$ $$ x\cdot(x^2-3b)=0$$ $$x = 0, \pm\sqrt{3b}$$

Hence, we check $f(x)$ for $x\in\{-2,0,1,\pm\sqrt{3b}\}$ (If either one of $\pm\sqrt{3b}$ are not in the given range, we do not include them)

So, $$f(0)=b^2$$$$f(-2) = b^2 - 24b + 16$$$$f(1) = b^2 - 6b + 1$$$$f(\pm\sqrt{3b})= -8b^2$$

From this analysis, we know definitely that $b^2 \geq -8b^2$, so we remove $\pm\sqrt{3b}$ from contention.

So, out of the remaining three values of $x\in\{-2,0,1\}$, check which one leads to the largest value for $f(x)$.

Answer 2

Why don't you try to solve $$f'(x)=4x^3-12bx=0$$ for $$x$$?