Question

Determine the intervals on concavity for the function h(x) -

-x^4 + 2x^3 - 5x, and state the point(s) of inflection

Answer 1

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

Step-by-step explanation:

To determine the intervals of concavity for the function h(x) = -x^4 + 2x^3 - 5x, and to find the points of inflection, we must first find the second derivative of h(x). The concavity of a function is determined by the sign of its second derivative. A function is concave up where its second derivative is positive and concave down where its second derivative is negative. The points of inflection are points where the concavity changes, that is, where the second derivative is zero or undefined.

First, we find the first derivative h'(x):

1. h'(x) = -4x^3 + 6x^2 - 5.

Then we find the second derivative h''(x):

1. h''(x) = -12x^2 + 12x.

To find points of inflection, set h''(x) equal to zero and solve for x:

1. 0 = -12x^2 + 12x.
2. x(1-x) = 0, so x = 0 or x = 1.

To determine the concavity, look at the sign of h''(x) on intervals defined by x = 0 and x = 1:

• When x < 0, h''(x) is positive, so h(x) is concave up.
• When 0 < x < 1, h''(x) is negative, so h(x) is concave down.
• When x > 1, h''(x) is positive again, so h(x) is concave up.

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