The value 5 is an upper bound for the zeros of the function shown below.
f(x)=14+1'- 11x4 – 9x + 18

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.

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