Answer to Question #105052 in Calculus for Ogumo

Question

For what values of x is the graph of y=xe^−2x concave down?

Answer 1

Graph of function f(x) is concave down when f"(x) < 0

$y=xe^{-2x}$

Differentiating with respect to x

$y'=e^{-2x}+x\cdot e^{-2x}(-2)$

$\Rightarrow y'=e^{-2x}\left(1-2x\right)$

Differentiating with respect to x again

$y''=e^{-2x}(0-2)+(1-2x)e^{-2x}\cdot(-2)$

$\Rightarrow y''=e^{-2x}(-2-2(1-2x))$

$\Rightarrow y''=e^{-2x}(-2-2+4x)$

$\Rightarrow y''=(4x-4)e^{-2x}$

point of inflection when $y''=0$

Since $e^{-2x}$ is always positive

$(4x-4)=0$

$x=1$

We need to check sign of $y''$ for x < 1 and x >1

When x < 1 ( let x = 0)

$y'' = e^{-2 \cdot 0}(4 \cdot0-4)=1 (0-4)=-4 < 0$ ................concave downward

when x > 1 (let x=2)

$y''=e^{-2 \cdot 2}(4 \cdot 2-4)=e^{-4}\cdot 4 \approx 0.07 >0$ ............concave upward

Thus, graph of $y=xe^{-2x}$ is concave down for x <1

In interval notation, graph of $y=xe^{-2x}$ is concave down in $(-\infty, 1)$