Question
\(f(x)=1-\sqrt[3]{x}\)
(A) Find the intervals on which f is increasing or decreasing.
(B) Locate all maxima and minima.
(C) Find the intervals over which f is concave upward or downward.
(D) Find all inflection points.
(E) Sketch the graph of f.
Answer/Explanation
(A)Use the first derivative to find where f is increasing or decreasing. The first derivative \(f'(x)=\frac{1}{3}x^{-\frac{2}{3}}< 0\) for (-∞, +∞), then f is decreasing for all values of x.
(B) First, find any points where f ′(x) = 0, but f ′ < 0 for all values of x. At x = 0, f ′(x) does not exist since there is a zero in the denominator. Therefore, there are no maxima or minima.
(C) Use the second derivative to test for concavity.\)f”(x)=\frac{2}{9}x^{-\frac{5}{3}}\).f”>0 when x > 0; therefore, f is concave upward on the interval (0, +∞). If x < 0, then f ″ < 0; therefore, f is concave downward on (-∞, 0).
(D) The second derivative f ″ changes concavity at x = 0, so there is an inflection point at x = 0. The function f (0) = 1, so the inflection point is at coordinate (0, 1), which is also the y-intercept.
(E) y-intercept at (0, 1). x-intercept at (1, 0).
