Question

Let f(x) = x^1/2(x - 2), x ∈ R & x > 0. Then the critical point of f is

Answer 1

f(x)= \(x^\frac12\)(x - 2), x ∈ R & x > 0 .

Domain of  f(x) is (0 , 0).

∴ f¹(x) = \(x^\frac12\) + \(\frac12\)\(x^\frac{-1}2\)(x - 2)

for critical point, we have f¹(x) = 0

∴ \(x^\frac12\) + \(\frac12\)\(x^\frac{-1}2\)(x - 2) = 0

\(\Rightarrow\) 2x + x - 2 = 0 ( on multiplying both sides by \(2x^\frac{1}2\))

\(\Rightarrow\) 3x - 2 = 0

\(\Rightarrow\) x = \(\frac23\)

Hence, x = \(\frac23\) is a critical point of f.