Find the critical numbers of the function f(x)=x^2 e^{-3x}.

Question:

Find the critical numbers of the function {eq}f(x)=x^2 e^{-3x}{/eq}.

Critical Numbers of a Function:

We need to understand the critical number of a function for solving this problem:

The values of x where {eq}f'(x) = 0 {/eq} or {eq}f'(x) {/eq} is undefined but {eq}f(x) {/eq} is defined that values are called the critical numbers of {eq}f(x) {/eq}.

Notes:

• Maximum and minimum values of a function occur only at critical numbers but a function may or may not have maximum/minimum values at a critical number.
• Critical number is also called critical value.

Given: {eq}f(x) = x^2 e^{- 3 x} {/eq}

We will compute the derivative of {eq}f(x) {/eq} for getting the critical number.

Take the derivative of {eq}f(x) {/eq} with respect to x

{eq}\begin{align*} \displaystyle f'(x) &= \frac{d}{dx} [ x^2 e^{- 3 x} ] \\ &= x^2 \frac{d}{dx} [e^{- 3 x} ] + e^{- 3 x} \frac{d}{dx}[x^2 ] &\text{(Applying the product rule for the derivative } \frac{d}{dx} [uv] = u \frac{dv}{dx} + v \frac{du}{dx} \text{)}\\ &= x^2 \times - 3 e^{- 3 x} + e^{- 3 x} \times 2 x &\text{(Differentiating using the formulas } \frac{d}{dx}[ e^{ax} ] = a e^{a x} \text{ and } \frac{d}{dx}[x^n] = n x^{n - 1} \text{)}\\ &= - 3 x^2 e^{- 3 x} + 2 x e^{- 3 x}\\ &= e^{- 3 x} ( 2 x - 3 x^2) \\ f'(x) &= 0 &\text{(Plugging in } f'(x) = 0 \text{ for getting the critical number)}\\ e^{- 3 x} ( 2 x - 3 x^2) &= 0 \\ 2 x - 3 x^2 &= 0 &\text{(Dividing both sides by } e^{- 3 x} \text{)}\\ x ( 2 - 3 x ) &= 0 \\ x &= 0 \text{ or } 2 - 3 x = 0 \\ x &= 0 \text{ or } x = \frac{2}{3} \end{align*} {/eq}

Therefore the critical numbers are {eq}\displaystyle x = \color{blue}{\frac{2}{3}} \text{ and } \color{blue}{0 } {/eq}.