Calculus with "a" Non-Smooth Function

The value of $a$ is irrelevant because extreme values are not affected by horizontal translation. Let's pick $a = 0$ for convenience and study $g(x) = 2|x|^2 - |x| + 3 = 2x^2 - |x| + 3$ .

Let's now search for critical points. We know that $|x|$ is not differentiable at $x=0$ so we should consider it. $g(0) = 3$ . Otherwise, we may restrict the function to the open set $\mathbb{R} - \{ 0 \}$ and study it normally, remembering that $|x|' = sgn(x)$ .

$g'(x) = 0 \\ 4x - sgn(x) = 0 \\ 4x = sgn(x) \\ 4|x| = 1 \\ |x| = \frac{1}{4}$

To go from the third to the fourth line I divided both sides by $sgn(x)$ . So the other critical points are $\frac{-1}{4}, \frac{1}{4}$ . Evaluating at these points (symmetry means we only need to check one) we get $f \left( \frac{1}{4} \right) = 3 - \frac{1}{8}$ which we know is smaller than $3$ so it is our minimum.

Note that $f(x) = |x-a| \ge 0$ is symmetrical about $x=a$ , where $f(a)=0$ . If we let $u=x-a$ , then $f(u) = |u|$ which is an even function. Then to consider the minimum of $g(x)$ , we need only to consider for $u \ge 0$ .

$\begin{aligned} g(x) & = 2f(x)^2 - f(x) + 3 \\ & = 2f(u)^2 - f(u) + 3 \\ & = 2|u|^2 - |u| + 3 & \small \color{#3D99F6} \text{Considering for }u \ge 0 \\ & = 2u^2 - u + 3 \\ & = 2\left(u-\frac 14\right) - \frac 18 + 3 \\ \implies \min (g(x)) & = 3 - \frac 18 = \boxed{2.875} & \small \color{#3D99F6} \text{when }u = \frac 14 \end{aligned}$

$g(x)= 2(\mid x-a \mid)^2$ $-$ $\mid x-a \mid$ $+$ $3$

$g'(x)$ $=$ $4\mid x-a \mid$ $\cdot$ $\frac{\mid x-a \mid}{x-a}$ $-$ $\frac{\mid x-a \mid}{x-a}$

Set $g'(x)=0$ and factor:

$0$ $=$ $\frac{\mid x-a \mid}{x-a}$ ( $4\mid x-a \mid$ $-1$ )

$1$ $=$ $4\mid x-a \mid$

$\frac{1}{4}$ $=$ $\mid x-a \mid$

This is only true if and only if

$x-a$ $=$ $\frac{1}{4}$

or

$x-a$ $=$ $-\frac{1}{4}$

In our case, both equations will be correct because $f(x)=\mid x-a \mid$ is a transformation of a linear function. This means if we undo the absolute value, then we will get $x-a$ which is a linear function defined for all x and y. This means there will be an instance where $x-a=-\frac{1}{4}$ and where $x-a=\frac{1}{4}$ .

Solving both expressions tells us that

$g'(x)=0$ if $x=a+ \frac{1}{4}$ or $x=a- \frac{1}{4}$

Note: $g'(x)$ is undefined at $x=a$

We can use the First Derivative Test to determine that $x=a-\frac{1}{4}$ and $x=a+\frac{1}{4}$ are minimums of $g(x)$ . We can plug both of these values into $g(x)$ and see which output is smaller.

$g(x)= 2(\mid x-a \mid)^2$ $-$ $\mid x-a \mid$ $+$ $3$

Note: $(\mid x-a \mid)^2 = (x-a)^2$

$g(a-\frac{1}{4})= 2(a-\frac{1}{4}-a)^2$ $-$ $\mid a-\frac{1}{4}-a \mid$ $+$ $3$

$=$ $2 \cdot (\frac{-1}{4})^2 - \mid -\frac{1}{4} \mid + 3$

$=$ $\frac{2}{16} -\frac{1}{4} + 3$

$=$ $\frac{2-4+48}{16}$

$=$ $\frac{46}{16}$

$=$ $2.875$

And if you plug $a+\frac{1}{4}$ into $g(x)$ , you will actually get the same answer.

Thus, the minimum possible output of $g(x)$ is $\boxed{2.875}$