The minimum value of the function f(x) = _0^2 e^|x-t|,dt is:

Question

The minimum value of the function $f(x) = \displaystyle\int_0^2 e^{|x-t|}\,dt$ is:

Accepted Answer

For $x\le0$: $f(x) = \displaystyle\int_0^2 e^{t-x}dt = e^{-x}(e^2-1)$ --- decreasing.

For $x\ge2$: $f(x) = \displaystyle\int_0^2 e^{x-t}dt = e^{x-2}(e^2-1)$ --- increasing.

For $0<x<2$: $f(x) = \displaystyle\int_0^x e^{x-t}dt+\int_x^2 e^{t-x}dt = e^x+e^{2-x}-2$.

Since $f$ decreases on $(-\infty,0]$ and increases on $[2,\infty)$, the minimum lies in $(0,2)$. By AM--GM, $e^x+e^{2-x}\ge 2\sqrt{e^x\cdot e^{2-x}}=2e$, with equality at $x=1$:

$f(1) = 2e-2 = 2(e-1)$

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