Answer to Question #274428 in Calculus for Artika

Question #274428

Find the relative extrema of 𝑓 π‘₯ = 3π‘₯^5 βˆ’ 5π‘₯^2.


1
Expert's answer
2021-12-06T16:47:51-0500

"\\text{Relative extrema can only occur at critical points}\\\\\n\\text{Given}, \\, f(x) = 3x^5 - 5x^2. \\, \\text{Now,}\\\\\nf'(x) = 15x^4 - 10x \\\\\n\\text{At critical points,} \\, f'(x) = 0 \\\\\n\\text{Hence}, 15x^4 - 10x = 0\\\\\n5x(3x^3 - 2) = 0 \\\\\n\\implies x = 0, \\, \\sqrt[3]{\\frac{2}{3}}\\\\\n\\text{Put these values into} \\, f(x) = 3x^5 - 5x^2. \\, \\text{Now},\\\\\nf(0) = 0 \\, \\, \\text{and}\\, \\, f(\\sqrt[3]{\\frac{2}{3}}) = -2.29.\\\\\n\\text{Hence,} \\, f(x) \\, \\text{has a local maximum at} \\, x = 0 \\\\\n\\text{and local minimum at} \\, x = \\sqrt[3]{\\frac{2}{3}}."


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