Final answer:
Using the Intermediate Value Theorem, we have shown that since f(x) is continuous and f(0) is -5 while f(2) is 5, it guarantees at least one zero for f(x) between x=0 and x=2.
Step-by-step explanation:
The Intermediate Value Theorem states that if a continuous function, f(x), takes on two values, f(a) and f(b), at any two points, a and b, then it also takes on any value between f(a) and f(b) at some point between a and b. To show that the given function f(x) = x² + 3x - 5 has a zero in the interval [0,2], we need to compute the values of the function at the endpoints of this interval.
For f(0) we have f(0) = 0² + 3(0) - 5 = -5, and for f(2) we find f(2) = 2² + 3(2) - 5 = 4 + 6 - 5 = 5. Since f(0) is negative and f(2) is positive, and the function f(x) is continuous, the Intermediate Value Theorem guarantees that f(x) has at least one zero between x = 0 and x = 2.