Question

For the function \(f\left(x\right)=x^2e^{-x}\) , the maximum occurs when x is equal to

Answer 1

f(x) will be maximum at x = 2.

Step-by-step explanation:

The given function is $f(x) = x^{2} e^{- x}$ ........... (1)

Now, we have to find the value of x when f(x) is maximum.

Then the condition for maximum f(x) is $\frac{df(x)}{dx} = 0$.

So, differentiating equation (1) with respect to x both sides we get,

$\frac{df(x)}{dx} = f'(x) = 2xe^{- x} - x^{2}e^{-x} = 0$

⇒ x(2 - x) = 0 {Since $e^{- x} \neq 0$ }

⇒ x = 0 or x = 2

Now, f(0) = 0 and f(2) = 0.54 {From equation (1)}

So, f(x) will be maximum at x = 2. (Answer)