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Question: Let \(y = - {x^2}{e^x}\), then the interval in which y increases with respect to x is: A) \(\left...

Let y=−x2exy = - {x^2}{e^x}, then the interval in which y increases with respect to x is:
A) (−2,0)\left( { - 2,0} \right)
B) (−inf⁡,−2)\left( { - \inf , - 2} \right)
C) (0,inf⁡)\left( {0,\inf } \right)
D) None of these

Explanation

Solution

We will first differentiate the function to find whether the function is increasing or decreasing in a given domain. Once we obtain the trend of the function we can use the given condition to find which one is the suitable option for the question. We will use the product rule of differentiation in order to find the derivative of the function.

Complete step by step solution:
We have, y=−x2exy = - {x^2}{e^x}
Differentiating both sides with respect to x, we get,
⇒dydx=ddx(−x2ex)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( { - {x^2}{e^x}} \right)
Taking negative sign out of the differential, we get,
⇒dydx=−ddx(x2ex)\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{d}{{dx}}\left( {{x^2}{e^x}} \right)
Applying the product rule of differentiation d(f(x)×g(x))dx=f(x)g′(x)+g(x)f′(x)\dfrac{{d\left( {f\left( x \right) \times g\left( x \right)} \right)}}{{dx}} = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right), we get,
⇒dydx=−[(x2)ddx(ex)+(ex)ddx(x2)]\Rightarrow \dfrac{{dy}}{{dx}} = - \left[ {\left( {{x^2}} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right) + \left( {{e^x}} \right)\dfrac{d}{{dx}}\left( {{x^2}} \right)} \right]
Using the power rule of differentiation as d(xn)dx=nxn−1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}, we get,
⇒dydx=−[(x2)(ex)+(ex)(2x)]\Rightarrow \dfrac{{dy}}{{dx}} = - \left[ {\left( {{x^2}} \right)\left( {{e^x}} \right) + \left( {{e^x}} \right)\left( {2x} \right)} \right]
Taking common terms outside the bracket, we get,
⇒dydx=−xex(x+2)\Rightarrow \dfrac{{dy}}{{dx}} = - x{e^x}\left( {x + 2} \right)
Since the function has to be increasing. So, we have, dydx>0\dfrac{{dy}}{{dx}} > 0.
Now, dydx=−xex(x+2)>0\dfrac{{dy}}{{dx}} = - x{e^x}\left( {x + 2} \right) > 0
Shifting the negative sign and changing the inequality sign, we get,
⇒xex(x+2)<0\Rightarrow x{e^x}\left( {x + 2} \right) < 0
Now, we know that exponential function is always positive. So, we get,
⇒x(x+2)<0\Rightarrow x\left( {x + 2} \right) < 0
Now, if the product of two expressions is negative. Then, one of them is positive and the other is negative.
So, either (x+2)<0\left( {x + 2} \right) < 0 and x>0x > 0, or x<0x < 0 and x+2>0x + 2 > 0.
If (x+2)<0\left( {x + 2} \right) < 0 and x>0x > 0, we get,
⇒x<\-2\Rightarrow x < \- 2 and ⇒x>0 \Rightarrow x > 0
Since these two inequalities have nothing in common. So, it yields no solution.
If x<0x < 0 and (x+2)>0\left( {x + 2} \right) > 0, we get,
⇒x<0\Rightarrow x < 0 and ⇒x>−2 \Rightarrow x > - 2
So, x∈(−2,0)x \in \left( { - 2,0} \right).
Therefore, the interval in which y increases with respect to x where y=−x2exy = - {x^2}{e^x} is (−2,0)\left( { - 2,0} \right). So, the correct option is A.

Note:
We must keep one thing in mind that, when we assume a function, it must be continuous as well as differentiable, to satisfy the fundamental conditions of differentiability. Transposition method is used to solve algebraic equations easily. In case of algebraic inequality, we have to change the sign of inequality by shifting the sign in it.