Question

If $f\left(x\right)=\underset{x-1}{\overset{x+1}{\int }}{e}^{-(t-1{)}^2}dt$, then the maximum value of $f\left(x\right)$ will occur at $x$ equal to
(correct answer + 2, wrong answer - 0.50)

• A. $2$
• B. $\sqrt{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

$f\left(x\right)=\underset{x-1}{\overset{x+1}{\int }}{e}^{-(t-1{)}^2}dt\Rightarrow f^{\prime }\left(x\right)=e^{-(x+1-1{)}^2}-e^{-(x-1-1{)}^2}\Rightarrow f^{\prime }\left(x\right)=e^{-(x{)}^2}-e^{-(x-2{)}^2}$
For critical point of the function,
$f^{\prime }\left(x\right)=0\Rightarrow 0=e^{-(x{)}^2}-e^{-(x-2{)}^2}\Rightarrow -(x{)}^2=-(x-2{)}^2\Rightarrow \pm x=(x-2)\Rightarrow -x=x-2\Rightarrow 2x=2\Rightarrow x=1$
Checking for maxima and minima,
$f^{\prime \prime }\left(x\right)=-2x\times e^{-(x{)}^2}+2(x-2)e^{-(x-2{)}^2}\Rightarrow f^{\prime \prime }\left(1\right)=-2e^{-1}-2e^{-1}=-4e^{-1}<0$

Hence, at $x=1$ the function will have maxima.