Answer to Question #161314 in Calculus for King

Question #161314

Check the validity of Rolle’s Theorem for "f(x)=1+(x-2)^(1\/3)" on the interval [2, 10].


1
Expert's answer
2021-02-09T08:47:15-0500

"\\displaystyle\nf(x) = 1 + (x - 2)^{\\frac{1}{3}}\\\\\nf(2) = 1 + (2 - 2)^{\\frac{1}{3}} = 1 + 0 = 1\\\\\nf(10) = 1 + (10 - 2)^{\\frac{1}{3}} = 1 + 2 = 3\\\\\nf'(x) = \\frac{1}{3}(x - 2)^{-\\frac{2}{3}}\\\\\nf'(x) = 0\\,\\, \\implies \\frac{1}{3}(x - 2)^{-\\frac{2}{3}} = 0\\\\\n\n\\textsf{No solution exists} \\\\\n\\textsf{Therefore the function does not}\\\\\n\\textsf{satisfy Rolle's theorem because}\\,\\, f(2) \\neq f(10) \\\\\n\\textsf{and}\\,\\, f'(x) \\neq 0 \\,\\, \\textsf{for}\\,\\, 2 \\leq x \\leq 10"


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