Functions#11

$\begin{aligned} f(x) & = (x-2)(x-4)(x-6)(x-8) + 16 & \small \color{#3D99F6} \text{Let }u = x-5 \\ & = {\color{#3D99F6}(u+3)}{\color{#D61F06}(u+1)(u-1)}{\color{#3D99F6}(u-3)} + 16 \\ & = {\color{#3D99F6}(u^2-9)}{\color{#D61F06}(u^2-1)} + 16 \\ & = u^4-10u^2 + 9 + 16 \\ & = (u^2-5)^2 & \small \color{#3D99F6} \text{Note that }(u^2-5)^2 \ge 0 \\ & \ge \boxed{0} \end{aligned}$

I calculated the derivative as $f'(x)=4x^3-60x^2+280x-400$ . I set $f'(x)=0$ , and proceeded as follows:

$x^3-15x^2+70x-100=0$

$(x-5)(x^2-10x+20)=0$

I found $f(5)=25>16=f(2)$ , so $f(5)$ is not the absolute minimum.

I came up with a creative way to substitute the other two roots.

Suppose $a^2-10a+20=0$ .

$f(a)=(a-2)(a-8)(a-4)(a-6)$

$=(a^2-10a+16)(a^2-10a+24)$

$=(a^2-10a+20-4)(a^2-10a+20+4)$

$=(-4)(4)=16$ .

Since f is differentiable and possesses the end behavior nature that it does, the absolute minimum is $16$ .