Question

The value 5 is an upper bound for the zeros of the function shown below.

f(x)= x4 + x3 – 11.x2 – 9x + 18
A. True

B. False

Answer 1

Explanation:

A

Answer 2

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Explanation:

Given

`f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18`

We know the rational zeros theorem such as:

if
`x=c` is a zero of the function
`f(x)`,

then
`f(c) = 0`.

As the
`f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18` is a polynomial of degree
`4`, hence it can not have more than
`4` real zeros.

Let us put certain values in the function,

`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`

From the above calculation results, we determined that
`4` zeros as

`x = -3, -2, 1,` and
`3`.

Hence, we can check that

`f(x) = (x+3)(x+2)(x-1)(x-3)`

Observe that,

`for x > 3`,
`f(x)` increases rapidly, so there will be no zeros for
`x>3`.

Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.