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

For the functionf(x)=xexf(x)=x{{e}^{x}}the point
1)x=01)x=0 is a maximum
2)x=02)x=0 is a minimum
3)x=−13)x=-1 is a maximum
4)x=−14)x=-1 is a minimum

Explanation

Solution

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 xx now, double differentiate the function and put the value of xx 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)f'(x). Now equate the first derivative to00.By equating the first derivative to 00,we get the values ofxx. Again differentiate the first derivative to get the second derivative of the function. Now put the value of xxin the second derivative.
If the second derivative is less than zero then there is a local maximum at xxand if the second derivative is greater than zero then there is a local minimum.
The given function isf(x)=xexf(x)=x{{e}^{x}}differentiate the function to get the first derivative
f(x)=xexf(x)=x{{e}^{x}}
ddxf(x)=ddx(xex)\frac{d}{dx}f(x)=\frac{d}{dx}(x{{e}^{x}})
By using the product rule we get,
f′(x)=1×ex+x×exf'(x)=1\times {{e}^{x}}+x\times {{e}^{x}}
f′(x)=ex(x+1)f'(x)={{e}^{x}}(x+1)
For finding the maxima or minima equate the first derivative to00. So we get,
f′(x)=ex(x+1)=0f'(x)={{e}^{x}}(x+1)=0
From the above expression we can say that,
ex=0{{e}^{x}}=0or x+1=0x+1=0
Now solving for thexx, we get the value of xxas
x=−∞x=-\infty orx=−1x=-1
But,x=−∞x=-\infty is not possible practically so we discard this value. Now we have to further solve this question with the valuex=−1x=-1
Now differentiate the first derivative in order to get the second derivative

& f''(x)={{e}^{x}}+1\times {{e}^{x}}+x\times {{e}^{x}} \\\ & f''(x)=2{{e}^{x}}+x{{e}^{x}} \\\ \end{aligned}$$ Now put the value of$$x=-1$$,we get $$\begin{aligned} & f''(-1)=2{{e}^{-1}}+(-1){{e}^{-1}} \\\ & f''(-1)=\frac{2}{e}-\frac{1}{e} \\\ & f''(-1)=\frac{1}{e} \\\ \end{aligned}$$ We know that the approximate value of$$e$$is$$2.718$$. So second derivative becomes $$\begin{aligned} & f''(x)=\frac{1}{2.718} \\\ & f''(x)=0.3679 \\\ \end{aligned}$$ We observe that the second derivative is greater than$$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.