Question

On which of the following intervals does the intermediate value theorem guarantee that f (x) has a zero? Select all correct answers. f(x) = 3x² – 1

a) (-5,3)
b) (-1,0)
c) (0,4)
d) None of these intervals.

Answer 1

The Intermediate Value Theorem guarantees that the function f(x) = 3x² − 1 has a zero in both intervals (-1,0) and (0,4), due to the change in sign of f(x) from positive to negative and vice versa.

Step-by-step explanation:

You're asking about the Intermediate Value Theorem and where the function f(x) = 3x² − 1 has a zero. To answer this, we need to consider the intervals where the function changes from a positive to a negative value or vice versa, indicating that it must cross the x-axis (where f(x) = 0).

For interval (-5,3), at x = -5, f(x) = 74, and at x = 3, f(x) = 26, both of which are positive. So there is no guarantee of a zero here.

Interval (-1,0) has f(-1) = 2 and f(0) = -1. Here, we have a sign change from positive to negative, so according to the Intermediate Value Theorem, there must be a zero in this interval.

Looking at interval (0,4), we have f(0) = -1 and f(4) = 47, another sign change from negative to positive, which means there is also a zero in this interval.

Therefore, the answers are intervals (-1,0) and (0,4).