Question

Find the critical numbers of the function.

35.\(f(x) = {x^2}{e^{ - 3x}}\).

Short answer

The critical numbers of the function \(f(x)\) are \(x = 0,\;x = \frac{2}{3}\).

Step-by-step solution

Given data

The function is \(f(x) = {x^2}{e^{ - 3x}}\).

Concept of critical number

A critical number of a function\(f\)is a number\(c\), if it satisfies either of the below conditions:

(1)\({f^\prime }(c) = 0\)

(2)\({f^\prime }(c)\), Does not exist.

Obtain the first derivative of the given function

Obtain the first derivative of the given function.

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{x^2}{e^{ - 3x}}} \right)\\ &= {x^2}\frac{d}{{dx}}\left( {{e^{ - 3x}}} \right) + {e^{ - 3x}}\frac{d}{{dx}}\left( {{x^2}} \right)\\ &= {x^2}\left( { - 3{e^{ - 3x}}} \right) + {e^{ - 3x}}(2x)\\ &= x{e^{ - 3x}}( - 3x + 2)\end{aligned}\)

Obtain the critical numbers

Take \({f^\prime }(x) = 0\)and obtain the critical numbers.

\(\begin{aligned}{c}x{e^{ - 3x}}( - 3x + 2) &= 0\\x = 0,\;2 - 3x &= 0\\x &= 0,\;x &= \frac{2}{3}\end{aligned}\)

Thus, the solutions are numbers are \(0,\;\frac{2}{3}\).

The critical numbers of the function \(f(x)\) are \(x = 0,\;x = \frac{2}{3}\).