Question

Find the relative extrema of 𝑓 𝑥 = 3𝑥^5 − 5𝑥^2.

Answer 1

$\text{Relative extrema can only occur at critical points}\\ \text{Given}, \, f(x) = 3x^5 - 5x^2. \, \text{Now,}\\ f'(x) = 15x^4 - 10x \\ \text{At critical points,} \, f'(x) = 0 \\ \text{Hence}, 15x^4 - 10x = 0\\ 5x(3x^3 - 2) = 0 \\ \implies x = 0, \, \sqrt[3]{\frac{2}{3}}\\ \text{Put these values into} \, f(x) = 3x^5 - 5x^2. \, \text{Now},\\ f(0) = 0 \, \, \text{and}\, \, f(\sqrt[3]{\frac{2}{3}}) = -2.29.\\ \text{Hence,} \, f(x) \, \text{has a local maximum at} \, x = 0 \\ \text{and local minimum at} \, x = \sqrt[3]{\frac{2}{3}}.$