Question

How do you find the local extrema for $f(x)=x{{e}^{x}}$?

Answer 1

From the given question we have to find the local extrema of $f(x)=x{{e}^{x}}$. As we know that if ${{x}_{0}}$ is a turning point of the function f then ${{f}^{'}}({{x}_{0}})=0$. Therefore, the return points of the function f will be among the solutions of the equation ${{f}^{'}}(x)=0$. So, we will equate the derivative of the function to zero and then we will search among the solutions in which the derivative has a change of sign. If the derivative is positive, we know that the function is increasing, whereas if the derivative is negative, then the function is decreasing. When the derivative changes from negative to positive, the function has a local minimum, whereas if the change of sign is reversed, that is, from positive to negative, then the function has a local maximum.

Complete step by step answer:
From the given question we have to find the local extrema of
$\Rightarrow f(x)=x{{e}^{x}}$
As we know that if ${{x}_{0}}$ is a turning point of the function f then
$\Rightarrow {{f}^{'}}({{x}_{0}})=0$
Therefore, the return points of the function f will be among the solutions of the equation
$\Rightarrow {{f}^{'}}(x)=0$
So, we will equate the derivative of the function to zero and then we will search among the solutions in which the derivative has a change of sign. If the derivative is positive, we know that the function is increasing, whereas if the derivative is negative, then the function is decreasing. When the derivative changes from negative to positive, the function has a local minimum, whereas if the change of sign is reversed, that is, from positive to negative, then the function has a local maximum.
The function is
$\Rightarrow f(x)=x{{e}^{x}}$
By differentiating the function, we will get,
$\Rightarrow {{f}^{'}}(x)=e^x+x\times {{e}^{x}}$
By equating to zero we will get,
$\Rightarrow \left( 1+x \right)\times {{e}^{x}}=0$
$\Rightarrow 1+x=0$
$\Rightarrow x=-1$
Now we find the double derivative of $f\left( x \right)$. So, we get,
$\Rightarrow {{f}^{'}}(x)=e^x+x\times {{e}^{x}}$
$\Rightarrow f''(x)={{e}^{x}}+{{e}^{x}}\left( x+1 \right)$
$\Rightarrow f''(x)={{e}^{x}}\left( x+2 \right)$
We got the critical point in the above as $\Rightarrow x=-1$. Now, we will substitute this in double derivative function.
$\Rightarrow f''(-1)={{e}^{-1}}\left( 1 \right)=\dfrac{1}{e}>0$
Hence, we can say it as the local minimum.
It is easy to verify that, for values of $x<-1$, the derivative is negative, ${{f}^{'}}(x)<0$, while for values of $x>-1$, the derivative is positive, ${{f}^{'}}(x)>0$. This means that $x=-1$ is a relative minimum.
Now, the y coordinate is obtained by replacing the value of x in the equation of the function.
Then we will get,
$\Rightarrow y=\left( -1 \right){{e}^{-1}}$
$\Rightarrow y=\dfrac{1}{e}$

Therefore, the relative minimum at the point is $\left( -1,\dfrac{1}{e} \right)$

Note: Students should know the concept of function. Students should know when there is a local maximum and when there is a local minimum. Students should not be confused in the concept - if the derivative is positive, we know that the function is increasing, whereas if the derivative is negative, then the function is decreasing. We should note that when the derivative changes from negative to positive, the function has a local minimum, whereas if the change of sign is reversed, that is, from positive to negative, then the function has a local maximum.