If f(x) is an increasing function and g(x) is a decreasing function such that gof(x) is defined, then gof(x) will be

Question

Question

This question was previously asked in

Airforce Group X MBT 15-Jul-2021 Shift 1

Answer 1

Concept:

Calculation:

Given: f(x) is an increasing function and g(x) is a decreasing function

∴ f’(x) > 0 and g’(x) < 0

Let h(x) = gof(x) = g(f(x))

Differentiating with respect to x, we get

⇒ h’(x) = g’(f(x)) × f’(x)

We know, f’(x) > 0 and g’(x) < 0

Therefore, h’(x) = (Negative) × (Positive) = Negative

∴ h’(x) < 0

Hence, gof(x) is decreasing function.

Alternate solution:

Let x₁ < x₂

Given: f(x) is an increasing function and g(x) is a decreasing function

∴ f(x₁) < f(x₂) and g(x₁) > g(x₂)

Let h(x) = gof(x) = g(f(x))

We know that, f(x₁) < f(x₂)

⇒ g(f(x₁)) > g(f(x₂))

Hence, gof(x) is decreasing function.