The interval in which the function f(x)=xˣ, x>0, is strictly increasing is

Solution

Maths sir · 4 steps

Write xˣ as eˣˡⁿˣ, differentiate, and the sign of 1+ln x decides everything.

1. Rewrite the power. Both the base and the exponent contain x, so neither the power rule nor the exponential rule applies directly. For x>0, write f(x) = xˣ = eˣˡⁿˣ.

   f'(x) = eˣˡⁿˣ(ln x + 1) = xˣ(1+ln x)

2. Check the sign. The factor xˣ is always positive for x>0, so the sign of f' is the sign of 1+ln x. This is positive when ln x > -1, that is when x > 1/e, and negative for 0 < x < 1/e.
3. The endpoint. At x = 1/e we get f' = 0, but only at that single point. A zero derivative at one isolated point does not stop a function from being strictly increasing, so 1/e can be included.
4. So. f(x) = xˣ is strictly increasing on [1/e, ∞), which is option (d).

The idea to keep. For a variable raised to a variable power, write it as eᵍ⁽ˣ⁾ˡⁿᶠ⁽ˣ⁾ before differentiating. A derivative that is zero only at isolated points still allows a strictly increasing function.

Answer from the NTA final answer key. Solution by PrepWiser.