Question

For the function $f\left(x\right)=x{e}^x$ the point

• A. $x=0$is a maximum
• B. $x=0$is a minimum
• C. $x=-1$is a maximum
• D. $x=-1$is a minimum

Answer 1

The correct option is D

$x=-1$is a minimum

Explanation for correct option

Step1.First order diffrentiation

Differentiate with respect to $x$.

$\begin{array}{rcl}\Rightarrow f\left(x\right)& =& x{e}^x\\ \Rightarrow f\text{'}\left(x\right)& =& 1\times e^x+x{e}^x\\ f\text{'}\left(x\right)& =& 0\\ & \Rightarrow & e^x+x{e}^x=0\\ \Rightarrow e^x(1+x)& =& 0\\ e^x& =& 0or1+x=0\\ x& =& \infty or-1\end{array}$

Step2. Double Differentiation

Differentiate $f\text{'}\left(x\right)=1\times e^x+x{e}^x$again

$\begin{array}{rcl}\Rightarrow f\text{'}\left(x\right)& =& e^x+x{e}^x\\ \Rightarrow f\text{'}\text{'}\left(x\right)& =& e^x+e^x+x{e}^x\\ \Rightarrow f\text{'}\text{'}\left(x\right)& =& 2e^x+x{e}^x\\ \Rightarrow f\text{'}\text{'}(-1)& =& 2e^{-1}-e^{-1}\mathbf{\text{[x=-1]}}\\ \Rightarrow f\text{'}\text{'}(-1)& =& e^{-1}\\ \Rightarrow f\text{'}\text{'}(-1)& =& \frac{1}{e}>0\end{array}$

$\begin{array}{rcl}x& =& -1\end{array}$is the point of minimum
Hence, the correct option is option (D).