Question

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

Answer 1

Answer:

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.

Step-by-step 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.