Question

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

Answer 1

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