Question

The curve $y=x{{e}^{x}}$ has minimum value equal to
$\left( a \right)\text{ }\dfrac{-1}{e}$
$\left( b \right)\text{ }\dfrac{1}{e}$
(c) – e
(d) e

Answer 1

Hint : At first, differentiate the function and equate it to zero to know the points or the values of 0 we have to check on. Then find the second derivative and check the values for which the first derivative is 0. The value of x for which the second derivative is positive, it refers to the point of minima and if negative, it refers to the point of maxima.

Complete step-by-step answer :
In this question, we are given a curve $y=x{{e}^{x}}$ and we have to find the minimum value of it. Now, to find the minimum value y, we will first differentiate it and then equate it to 0. Then we will check those points only for the minimum. If the value at the point is the lowest, then that point is considered as a point of minima and the value is the minimum value. Also, one should check for minimum points but the points are not given, so we should write it here.
Now, we know that, $y=x{{e}^{x}}$
If a function is given is in the form of ${{e}^{x}}f\left( x \right)$ then its value after differentiation would be ${{e}^{x}}\left( f\left( x \right)+{{f}^{'}}\left( x \right) \right).$
Here, f (x) = x, so the value of f’(x) = 1.
So according to the formula, the value of $x{{e}^{x}}$ after differentiation will be ${{e}^{x}}\left( x+1 \right)$ which is represented as ${{y}_{1}}.$
${{y}_{1}}={{e}^{x}}\left( x+1 \right)$
Now, we will find the values of x at which ${{y}_{1}}$ will be equal to 0. So, we can say that,
${{y}_{1}}=0$
$\Rightarrow {{e}^{x}}\left( x+1 \right)=0$
Now, as we know that ${{e}^{x}}$ is an exponential function, so it can’t be 0. Hence,
x + 1 = 0
Therefore, x = – 1
Now, we will check for the second derivative of y whether it is positive or negative for the point at x = – 1. If the second derivative at x = – 1 is negative, then the point is a point of maxima and if it is positive, then the point is a point of minima. So, we know that,
${{y}_{1}}={{e}^{x}}\left( x+1 \right)$
Now, again using the identity that if we differentiate ${{e}^{x}}f\left( x \right)$ then we get,
${{e}^{x}}\left( f\left( x \right)+{{f}^{'}}\left( x \right) \right)$
Here, f(x) is (x + 1), then f’(x) is 1. So, then ${{y}_{2}}$ will be
${{y}_{2}}={{e}^{x}}\left( x+1+1 \right)$
$\Rightarrow {{y}_{2}}={{e}^{x}}\left( x+2 \right)$
Here, ${{y}_{2}}$ represents the double differentiation of y.
At x = – 1, the value of ${{y}_{2}}$ will be,
${{y}_{2}}={{e}^{-1}}\left( -1+2 \right)$
$\Rightarrow {{y}_{2}}=\dfrac{1}{e}\left( -1+2 \right)$
Here, $\dfrac{1}{e}$ is positive, hence it is of minimum value.
So, at x = – 1, the value of y will be
$y=x{{e}^{x}}$
$\Rightarrow y=\left( -1 \right){{e}^{-1}}$
$\Rightarrow y=\dfrac{-1}{e}$
Hence, the minimum value is $\dfrac{-1}{e}.$
Therefore, the correct option is (a).

Note : If the second derivative is 0 then this means it’s a point of inflection which means at this point the second derivative will change its sign.