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Question: Find the interval in which \(y={{x}^{2}}{{e}^{-x}}\) is an increasing function....

Find the interval in which y=x2e−xy={{x}^{2}}{{e}^{-x}} is an increasing function.

Explanation

Solution

We need to find the derivative form of the given function y=x2e−xy={{x}^{2}}{{e}^{-x}} to find f′(x)>0{{f}^{'}}\left( x \right)>0.
The interval in which the inequation satisfies f′(x)>0{{f}^{'}}\left( x \right)>0 will be the interval in which y=x2e−xy={{x}^{2}}{{e}^{-x}} is an increasing function. We find the derivation with the help of by product rule. We find the solution of the inequation to solve the problem.

Complete step by step answer:
The given function is y=x2e−xy={{x}^{2}}{{e}^{-x}}. The function will be an increasing function if we find that f′(x)>0{{f}^{'}}\left( x \right)>0. This means the second derivative of the curve will have to be positive.
We find the interval in which f′(x)>0{{f}^{'}}\left( x \right)>0 happens.
We take the derivative of the function y=x2e−xy={{x}^{2}}{{e}^{-x}} by differentiating both sides with respect to x. So, f′(x)=d(y)dx=d(x2e−x)dx{{f}^{'}}\left( x \right)=\dfrac{d\left( y \right)}{dx}=\dfrac{d\left( {{x}^{2}}{{e}^{-x}} \right)}{dx}.
We use the formula of by products we solve the equation
f′(x)=e−xd(x2)dx+x2d(e−x)dx=2xe−x−e−xx2{{f}^{'}}\left( x \right)={{e}^{-x}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+{{x}^{2}}\dfrac{d\left( {{e}^{-x}} \right)}{dx}=2x{{e}^{-x}}-{{e}^{-x}}{{x}^{2}}
We need to find out the interval of x for which f′(x)>0{{f}^{'}}\left( x \right)>0.
This means 2xe−x−e−xx2>02x{{e}^{-x}}-{{e}^{-x}}{{x}^{2}}>0. We know that for any values of x, e−x{{e}^{-x}} will be positive.
2xe−x−e−xx2>0 ⇒e−x(2x−x2)>0 ⇒(2x−x2)>0 \begin{aligned} & 2x{{e}^{-x}}-{{e}^{-x}}{{x}^{2}}>0 \\\ & \Rightarrow {{e}^{-x}}\left( 2x-{{x}^{2}} \right)>0 \\\ & \Rightarrow \left( 2x-{{x}^{2}} \right)>0 \\\ \end{aligned}
Now we solve quadratic inequation of x to find the interval.
(2x−x2)>0 ⇒(x2−2x)<0 ⇒x(x−2)<0 \begin{aligned} & \left( 2x-{{x}^{2}} \right)>0 \\\ & \Rightarrow \left( {{x}^{2}}-2x \right)<0 \\\ & \Rightarrow x\left( x-2 \right)<0 \\\ \end{aligned}
From the factorisation we break the interval region as x∈(0,2)x\in \left( 0,2 \right).

So, the interval in which y=x2e−xy={{x}^{2}}{{e}^{-x}} is an increasing function is (0,2)\left( 0,2 \right).

Note: We need to remember the function becomes increasing when the slope of the curve is greater than 0. This means for every increasing value of x; we find an increasing value of f(x)f\left( x \right). The value of e−x{{e}^{-x}} is always positive as we are taking the power value of a positive and that can never be changed into negative unless we change the base value. That’s why we took out e−x{{e}^{-x}} part of the equation in the first step.