Question

Find the smallest integer a such that the Intermediate Value Theorem guarantees that f(x) has a zero on the interval [0,a].

f(x)=3x²−8x−7

Answer 1

The smallest integer a such that f(x) has a zero on the interval [0, a] is a = 2.

Step-by-step explanation:

The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b] and takes on values of opposite sign at the endpoints, then it must have a zero between a and b.

In this case, we have f(x) = 3x² - 8x - 7. To find the smallest integer a such that f(x) has a zero on the interval [0, a], we need to determine the value of a.

Plugging in a = 0, we have f(0) = 3(0)² - 8(0) - 7 = -7. Since f(0) is negative, we need to find the smallest positive integer a such that f(a) is positive.

Let's try a = 1: f(1) = 3(1)² - 8(1) - 7 = -12. Since f(1) is still negative, we try a = 2: f(2) = 3(2)² - 8(2) - 7 = 1. Finally, f(2) is positive, so the smallest integer a such that f(x) has a zero on the interval [0, a] is a = 2.