Mathematics - Application of Derivative Question with Solution | TestHub

Given $f(x)=x{e}^{-a{x}^2}$. We find the first derivative: $f^{\prime }(x)=e^{-a{x}^2}+x(-2ax)e^{-a{x}^2}=e^{-a{x}^2}(1-2a{x}^2)$. Since $f(x)$ has a local extremum at $x=1$, we must have $f^{\prime }(1)=0$.$e^{-a}(1-2a)=0⟹1-2a=0⟹a=1/2$. So, the function is $f(x)=x{e}^{-x^2/2}$.

To find the maximum value in $[0,2]$, we find critical points by setting $f^{\prime }(x)=0$: $e^{-x^2/2}(1-x^2)=0⟹1-x^2=0⟹x=\pm 1$. The critical point in the interval $[0,2]$ is $x=1$. We evaluate $f(x)$ at the critical point and the endpoints of the interval:$f(0)=0\cdot e^0=0$.$f(1)=1\cdot e^{-1^2/2}=e^{-1/2}$.$f(2)=2\cdot e^{-2^2/2}=2e^{-2}$. Comparing the values: $f(0)=0$, $f(1)\approx 0.606$, $f(2)\approx 0.271$. The maximum value $M=e^{-1/2}$. The required value is $e{M}^2=e(e^{-1/2}{)}^2=e\cdot e^{-1}=1$. Rounded to two decimal places, the answer is $1.00$.