Question

For the curve $y=x{{e}^{x}}$ , the point
(a) $x=-1$ is minimum.
(b) $x=0$ is minimum.
(c) $x=-1$ is maximum.
(d) $x=0$ is maximum.

Answer 1

Hint: Differentiate the function $y=x{{e}^{x}}$ to equate it to zero, to calculate the critical points. Then, we need to differentiate the function once again to calculate $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ . If at the critical point, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}<0$ then we will have the maximum value of function at that point. If at the critical point, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}>0$ then we will have the minimum value of function at that point. If at the critical point, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$ then we will have neither maximum nor minimum value at that point.

Complete step by step solution:
We know that for any curve $y=f\left( x \right)$ , the maxima and minima points, and the points of inflection are known as critical points. To find these critical points, we differentiate the function and equate it to 0, so that, $\dfrac{dy}{dx}=0$ .
Further, if at that point $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}<0$ , the point is called a maxima, and the function is maximum,
If at that point $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}>0$ , the point is called a minima, and the function is minimum, and
If at that point $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$ , the point is called the point of inflection.
In our problem, we have $y=x{{e}^{x}}$ .
On differentiating this function, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( x{{e}^{x}} \right)$
For the RHS, we have to use Product Rule $\dfrac{d\left( u\cdot v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ .
So, we can write
$\dfrac{dy}{dx}=x\dfrac{d\left( {{e}^{x}} \right)}{dx}+{{e}^{x}}\dfrac{d\left( x \right)}{dx}$
$\Rightarrow \dfrac{dy}{dx}=x{{e}^{x}}+{{e}^{x}}...\left( i \right)$
$\Rightarrow \dfrac{dy}{dx}={{e}^{x}}\left( x+1 \right)$
Now, for critical points we must have
$\dfrac{dy}{dx}=0$
Putting the value of $\dfrac{dy}{dx}$ , we get
${{e}^{x}}\left( x+1 \right)=0$
So, either ${{e}^{x}}=0$ or $\left( x+1 \right)=0$ .
We know that for $x\in \left( -\infty ,\infty \right),\text{ }{{e}^{x}}>0$ . So, ${{e}^{x}}\ne 0$ .
$\begin{align}
  & \therefore \left( x+1 \right)=0 \\
 & \Rightarrow x=-1 \\
\end{align}$
So, $x=-1$ is a critical point.
Now, we have to calculate $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ , to determine which type of critical pint it is.
So, differentiating (i), we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( x{{e}^{x}} \right)}{dx}+\dfrac{d\left( {{e}^{x}} \right)}{dx}$
Using the Product Rule again, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left\{ x\dfrac{d\left( {{e}^{x}} \right)}{dx}+{{e}^{x}} \right\}+{{e}^{x}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=x{{e}^{x}}+2{{e}^{x}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( x+2 \right)$
Putting $x=-1$ in this equation, we get
${{\left. \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=-1}}={{e}^{\left( -1 \right)}}\left( \left( -1 \right)+2 \right)$
$\Rightarrow {{\left. \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=-1}}=\dfrac{1}{e}$
Here, ${{\left. \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=-1}}>0$ .
So, the function will achieve its minimum value at $x=-1$ .

So, the correct answer is “Option A”.

Note: We should not be careless and should not skip steps, as an extra negative sign can entirely change the solution. We must note that we have to compare the double differentiation $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ only at the critical points and not at any general point.