Question

The function $f\left(x\right)=x(x+4)e^{-\frac{x}{2}}$ has its local maxima at $x=a$, then $a=$

• A. $2\sqrt{2}$
• B. $1-\sqrt{3}$
• C. $-1+\sqrt{3}$
• D. $-4$

Answer 1

The correct option is A $2\sqrt{2}$

The given equation is:

$f\left(x\right)=x(x+4)e^{-\frac{x}{2}}$

To find the extremum points we differentiate and equate it to zero

$\Rightarrow f^{\prime }\left(x\right)=(x+4)e^{-\frac{x}{2}}+x[e^{-\frac{x}{2}}-\frac{1}{2}(x+4\left)e^{-\frac{x}{2}}\right]$

$\Rightarrow f^{\prime }\left(x\right)=(x+4)e^{-\frac{x}{2}}+x{e}^{-\frac{x}{2}}-\frac{x(x+4)e^{-\frac{x}{2}}}{2}$

$f^{\prime }\left(x\right)=0$

$\Rightarrow (x+4)e^{-\frac{x}{2}}+x{e}^{-\frac{x}{2}}-\frac{x(x+4)e^{-\frac{x}{2}}}{2}=0$

$e^{-\frac{x}{2}}\ne 0$ as this is a exponential function and this requires exponent power to tend to infinity to make the exponential function reach zero. Hence

$\Rightarrow x+4+x-\frac{x^2}{2}-2x=0$

$\Rightarrow x^2=8$

$\Rightarrow x=2\sqrt{2}$

Now to find whether at the critical points we find a maxima or minima we use the second derivative test.

$\Rightarrow f^{\prime \prime }\left(x\right)=e^{-\frac{x}{2}}-\frac{1}{2}(x+4)e^{-\frac{x}{2}}+e^{-\frac{x}{2}}-\frac{1}{2}(x+4)e^{-\frac{x}{2}}+x(-\frac{1}{2}{e}^{-\frac{x}{2}}-\frac{1}{2}[e^{-\frac{x}{2}}-\frac{1}{2}(x+4)e^{-\frac{x}{2}}\left]\right)$

$\Rightarrow f^{\prime \prime }\left(x\right)=2e^{-\frac{x}{2}}-(x+4)e^{-\frac{x}{2}}-x{e}^{-\frac{x}{2}}+\frac{1}{4}x(x+4)e^{-\frac{x}{2}}$

$\Rightarrow f^{\prime \prime }\left(2\sqrt{2}\right)=2e^{-\sqrt{2}}-(2\sqrt{2}+4)e^{-\sqrt{2}}-2\sqrt{2}{e}^{-\sqrt{2}}+\frac{1}{4}2\sqrt{2}(2\sqrt{2}+4)e^{-\sqrt{2}}$

$\Rightarrow f^{\prime \prime }\left(2\sqrt{2}\right)=2e^{-\sqrt{2}}(1-\sqrt{2})+(4+2\sqrt{2})e^{-\sqrt{2}}(\frac{1}{\sqrt{2}}-1)$

$\Rightarrow 2e^{-\sqrt{2}}>0$

$\Rightarrow (1-\sqrt{2})<0$

$\Rightarrow (4+2\sqrt{2})e^{-\sqrt{2}}>0$

$\Rightarrow (\frac{1}{\sqrt{2}}-1)<0$

$\Rightarrow f^{\prime \prime }\left(2\sqrt{2}\right)<0$. Thus the point $x=2\sqrt{2}$ is a point of maxima. ....Answer