Question

For the function\[f(x)=x{{e}^{x}}\]the point
\[1)x=0\] is a maximum
\[2)x=0\] is a minimum
\[3)x=-1\] is a maximum
\[4)x=-1\] is a minimum

Answer 1

Hint : To get the solution of this question use the concept of maxima and minima. Firstly differentiate the given function to obtain the value of \[x\] now, double differentiate the function and put the value of \[x\] to know the maximum or minimum function. By using these steps you can find the answer.

Complete step-by-step solution:
In this question we have to find whether the function has maximum value or minimum value. Now firstly let us discuss the steps on how to define a function to have maximum value or minimum value at a certain point.
Firstly differentiate the given function to get the first derivative i.e.\[f'(x)\]. Now equate the first derivative to\[0\].By equating the first derivative to \[0\],we get the values of\[x\]. Again differentiate the first derivative to get the second derivative of the function. Now put the value of \[x\]in the second derivative.
If the second derivative is less than zero then there is a local maximum at \[x\]and if the second derivative is greater than zero then there is a local minimum.
The given function is\[f(x)=x{{e}^{x}}\]differentiate the function to get the first derivative
\[f(x)=x{{e}^{x}}\]
\[\frac{d}{dx}f(x)=\frac{d}{dx}(x{{e}^{x}})\]
By using the product rule we get,
\[f'(x)=1\times {{e}^{x}}+x\times {{e}^{x}}\]
\[f'(x)={{e}^{x}}(x+1)\]
For finding the maxima or minima equate the first derivative to\[0\]. So we get,
\[f'(x)={{e}^{x}}(x+1)=0\]
From the above expression we can say that,
\[{{e}^{x}}=0\]or \[x+1=0\]
Now solving for the\[x\], we get the value of \[x\]as
\[x=-\infty \]or\[x=-1\]
But,\[x=-\infty \]is not possible practically so we discard this value. Now we have to further solve this question with the value\[x=-1\]
Now differentiate the first derivative in order to get the second derivative
\[\begin{align}
  & f''(x)={{e}^{x}}+1\times {{e}^{x}}+x\times {{e}^{x}} \\
 & f''(x)=2{{e}^{x}}+x{{e}^{x}} \\
\end{align}\]
Now put the value of\[x=-1\],we get
\[\begin{align}
  & f''(-1)=2{{e}^{-1}}+(-1){{e}^{-1}} \\
 & f''(-1)=\frac{2}{e}-\frac{1}{e} \\
 & f''(-1)=\frac{1}{e} \\
\end{align}\]
We know that the approximate value of\[e\]is\[2.718\]. So second derivative becomes
\[\begin{align}
  & f''(x)=\frac{1}{2.718} \\
 & f''(x)=0.3679 \\
\end{align}\]
We observe that the second derivative is greater than\[0\]. So at\[x=-1\]is a minimum value.
So we can conclude that the option\[(4)\] is correct.

 Note: The maxima of a given function are the points on the graph at which if we slightly move left or right the value of the function decreases. Similarly, the minima of a given function are the points on the graph at which if we slightly move left or right the value of the function increases.