Question

the intervals on which the graph f is concave down and concave up

• A. concave up for $(-\infty ,2-\sqrt{2})\cup (2+\sqrt{2},\infty )$
• B. concave down in $(2-\sqrt{2},2+\sqrt{2})$
• C. concave up in $(2-\sqrt{2},2+\sqrt{2})$
• D. concave down for $(-\infty ,2-\sqrt{2})\cup (2+\sqrt{2},\infty )$

Answer 1

Answer: (A), (B)

concave up for $(-\infty ,2-\sqrt{2})\cup (2+\sqrt{2},\infty )$
concave down in $(2-\sqrt{2},2+\sqrt{2})$

$f\left(2\right)=f\left(0\right)=0$
Since k>0 and independent of $x$, we get
$f^{\prime }\left(x\right)=k{e}^{-x}\left[\right(2-2x)-(2x-x^2\left)\right]$
$=k{e}^{-x}[x^2-4x+2]$
Now
$f^{\prime }\left(x\right)>0$ implies
$x^2-4x+2>0$ since $k>0$ and $e^{-x}>0$ for all $x$.
$(x-2{)}^2-2>0$
Or
$(x-2{)}^2>2$
Or
$x-2>\sqrt{2}$ and $x-2<-\sqrt{2}$
Or
$x>2+\sqrt{2}$ and $x<2-\sqrt{2}$
Hence $f\left(x\right)$ is increasing in $xϵ(-\infty ,2-\sqrt{2})\cup (2+\sqrt{2},\infty )$ (cocave up) and decreasing in $(2-\sqrt{2},2+\sqrt{2})$ (concave down).