So I am asked to find the maximum and minimum of the function:
$$f(x)=(2+|x|)\sqrt{2+(2-x)^2}$$ on the interval $[-1,2]$
So I took the derivative and got:
$$\frac{ 2((x-2)|x|+x(x^2-3x+3)) }{|x|\sqrt{(2+(2-x)^2}}$$
which is in fact correct. So I set it equal to zero and try to find the critical points.
The problem is though, how do I solve for $2((x-2)|x|+x(x^2-3x+3))=0$??? I certainly can't see a way…
Best Answer
Note that $f$ is positive on the entire domain of consideration.
Note that $f$ is non increasing on $[-1,0]$ since it is the product of two non increasing, non negative functions. Hence we need only consider the points $\{-1,0\}$ in the context of $x \in [-1,0]$.
For $x\ge 0$, we have $f(x) = (x+2)\sqrt{x^2-4x+6}$, and $f'(x) = 2{(x-1)^2 \over \sqrt{x^2-4x+6} }$. In particular, $f'(x) \ge 0$ for $x \in [0,2]$. Hence we need only consider the points $\{0,2\}$.
Since $f$ is non increasing on $[-1,0]$ and increasing on $[0,2]$, the maximum value must occur at either $\{-1,2\}$, evaluating shows that the maximum occurs at $f(-1) = 3 \sqrt{11}$. Similarly, the minimum must occur at $f(0) = 2 \sqrt{6}$.