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find the interval in which function, f(x) = x2ex where x ∈ R, is strictly decreasing?

find the interval in which function, f(x) = x2ex where x ∈ R, is strictly decreasing? Correct Answer (-2, 0)

Concept:

  • If a function continuous on interval and differentiable in (a, b).

 If f’(x) > 0 for all x in (a, b), then f is increasing on .

 If f’(x) < 0 for all x in (a, b), then f is decreasing on .

 If f’(x) = 0 for all x in (a, b), then f is constant on .

 

Calculation:

Given that,

⇒ f(x) = x2ex where x ∈ (0, ∞)

First we check type of function in interval so,

Differentiate the function with respect to x, we get

⇒ f’(x) = 2x ex + x2 ex

⇒ f’(x) = x ex(2 + x)

for critical points put f’(x) = 0

⇒ x ex(2 + x) = 0

So, x = 0 and x = - 2

Draw number line with critical points

[ alt="F1 Aman.K 22-05-2020 Savita D2" src="//storage.googleapis.com/tb-img/production/20/05/F1_Aman.K_22-05-2020_Savita_D2.png" style="width: 209px; height: 83px;">

Note: we put alternate sign on number line starting from right side with positive sign.

From the number line we can say that

⇒ Interval (- ∞, - 2) ∪ (0, ∞) function f strictly increasing.

⇒ Interval (-2, 0) function f strictly decreasing.

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