Inflection Points and Intervals

[Qube]

Homework Statement

Suppose that a continuous function f(x) has horizontal tangent lines at x = -1, x = 0, and x = 1. If f"(x) = 60x^3 - 30x, then which of the following statements is/are true?

A) f(x) has a local max at x = 1
B) f(x) has a local min at x = -1
C) f(x) has an inflection point at x = 0

Homework Equations

Local maximums occur at critical points.

All points at which horizontal tangent lines occur are critical points because the existence of a horizontal tangent line at that point implies the existence of that point on the function, and as we know, critical points must exist in the domain of the function.

Therefore, x = 1, 0, and 1 are critical points.

We can use the second derivative test to test for local extrema.

The Attempt at a Solution

f"(-1) = - 30. x = -1 is a local max. B is true.

f"(1) = 30. x = 1 is a local min. A is true.

f(x) has an inflection point at x = 0; the second derivative is 0 at x = 0 and x = ±1/sqrt(2).

f"(x) changes sign around x = 0 from being positive in the interval (-1/sqrt(2), 0) and (0, 1/sqrt(2)).

Therefore, all three are true.

 

[Mark44]

Homework Statement

Suppose that a continuous function f(x) has horizontal tangent lines at x = -1, x = 0, and x = 1. If f"(x) = 60x^3 - 30x, then which of the following statements is/are true?

A) f(x) has a local max at x = 1
B) f(x) has a local min at x = -1
C) f(x) has an inflection point at x = 0

Homework Equations

Local maximums occur at critical points.

All points at which horizontal tangent lines occur are critical points because the existence of a horizontal tangent line at that point implies the existence of that point on the function, and as we know, critical points must exist in the domain of the function.

Therefore, x = 1, 0, and 1 are critical points.

We can use the second derivative test to test for local extrema.

The Attempt at a Solution

f"(-1) = - 30. x = -1 is a local max. B is true.

f"(1) = 30. x = 1 is a local min. A is true.

f(x) has an inflection point at x = 0; the second derivative is 0 at x = 0 and x = ±1/sqrt(2).

f"(x) changes sign around x = 0 from being positive in the interval (-1/sqrt(2), 0) and (0, 1/sqrt(2)).

Therefore, all three are true.

Your reasoning looks good to me.