Find the interval in which function f(x) = \(\frac{x}{9} + \frac{4}{x}\) where x ∈ R, is strictly decreasing?

Question

Question

Answer 1

Concept:

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

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

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

Calculation:

Given that,

⇒ f(x) = \(\frac{x}{9} + \frac{4}{x}\) where x ∈ R

To calculate the critical point

⇒ f’(x) = \(\frac{1}{9} - \frac{4}{{{x^2}}}\)

⇒ f’(x) = 0

⇒ \(\frac{1}{9} = \frac{4}{{{x^2}}}\)

⇒ x²= 36

⇒ x = ± 6

Draw number line with critical points

F1 Aman.K 22-05-2020 Savita D1

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

From the number line we can say that

⇒ Interval (- ∞, - 6] ∪ [6, ∞) function f is strictly increasing.

⇒ Interval [-6, 6] function f is strictly decreasing.

In our question interval given [-5, 2] is the right answer because it is subinterval of [-6, 6].