Answer: True
Step-by-step explanation:
According to the rational zeros theorem, if x=a is a zero of the function f(x), then f(a) = 0.
Given: f(x) = x⁴ + x³ - 11x² - 9x + 18
From the calculator, obtain
f(5) = 448
f(4) = 126
f(3) = 0
f(2) = -20
f(1) = 0
f(0) = 18
f(-1) = 16
f(-2) = 0
f(-3) = 0
The polynomial is of degree 4, so it has at most 4 real zeros.
From the calculations, we found all 4 zeros as x = -3, -2, 1, and 3.
Therefore
f(x) = (x+3)(x+2)(x-1)(x-3).
For x>3, f(x)increases rapidly. Therefore there are no zeros for x>3.
The statement that x=5 is an upper bound for the zeros of f(x) is true.