Question

Determine the intervals of concavity for the function h(x)=-x⁴+2 x³-5 x, and state the point(s) of inflection.

Answer 1

To find the intervals of concavity for h(x) = -x⁴ + 2x³ - 5x, we must calculate the second derivative, set it to zero to find critical points, and test intervals around those points. The function is concave downward on (-∞, 0) and (1, ∞), concave upward on (0, 1), and the points of inflection are (0, h(0)) and (1, h(1)).

Step-by-step explanation:

Determining Intervals of Concavity and Points of Inflection

To find the intervals of concavity and the point(s) of inflection for the function h(x) = -x⁴ + 2x³ - 5x, we first need to calculate the second derivative of h(x). The second derivative, h"(x), will tell us the concavity of the function: if h"(x) is greater than zero, the graph is concave upward, and if it is less than zero, the graph is concave downward. A change in sign of h"(x) indicates a point of inflection.

First, we find the first derivative, h'(x) = -4x³ + 6x² - 5. Then, we find the second derivative, h"(x) = -12x² + 12x. Setting h"(x) equal to zero gives us the potential points of inflection when x = 0 or x = 1.

To determine the concavity on the intervals defined by these points, we can test values from the intervals in h"(x). For any value of x between 0 and 1, such as 0.5, h"(0.5) is positive, indicating that the interval (0, 1) is concave upward. For values of x less than 0 or greater than 1, like -1 and 2 respectively, h"(-1) and h"(2) are both negative, indicating the intervals (-∞, 0) and (1, ∞) are concave downward. The points (0, h(0)) and (1, h(1)) are identified as the points of inflection since the sign of the second derivative changes at these values of x.

Therefore, the intervals of concavity are as follows: the curve is concave downward on (-∞, 0) and (1, ∞), while it is concave upward on (0, 1). The points of inflection are at (0, h(0)) and (1, h(1)).

Answer 2

The function h(x) = -x^4 + 2x^3 - 5x is concave up on (-∞, 0) and (1, ∞), concave down on (0, 1), with points of inflection at x=0 and x=1.

Step-by-step explanation:

To determine the intervals of concavity for the function h(x) = -x⁴ + 2x³ - 5x, we first need to find the second derivative, h''(x). The concavity of the function depends on the sign of the second derivative. Wherever h''(x) is positive, the function is concave up, and wherever h''(x) is negative, the function is concave down. A point of inflection occurs where the concavity changes, which happens where h''(x) = 0 or is undefined.

The first derivative is h'(x) = -4x³ + 6x² - 5, and the second derivative is h''(x) = -12x² + 12x. We find the critical points by setting the second derivative equal to zero: -12x² + 12x = 0, which simplifies to x(x - 1) = 0. Therefore, x = 0 and x = 1 are our critical points.

To determine the signs of the second derivative, we test values in the intervals x < 0, 0 < x < 1, and x > 1. Plugging these values into the second derivative:

For x < 0, use x = -1: h''(-1) > 0, so the interval is concave up.

For 0 < x < 1, use x = 0.5: h''(0.5) < 0, so the interval is concave down.

For x > 1, use x = 2: h''(2) > 0, so the interval is concave up.

Thus, the function h(x) is concave up on the intervals (-∞, 0) and (1, ∞), concave down on the interval (0, 1), and the points of inflection are at x=0 and x=1.