Chapter 2.3, Problem 47E

Solution

To graph the function $f\left(x\right)=x^4+3x^3-9x^2+2x+3$ and to approximate all of its zeroes.

The zeroes of the function $f\left(x\right)=x^4+3x^3-9x^2+2x+3$ are approximately:

  $x=-4.9$ , $x=-0.45$ , $x=1$ and $x=1.35$ .

Given:

  $f\left(x\right)=x^4+3x^3-9x^2+2x+3$ .

Concept Used:

Zeroes of a function:

Given a function $f\left(x\right)$ . A point $x$ is said to be a zero of $f$ if and only if $f\left(x\right)=0$ .

Now, $f\left(x\right)=0$ if and only if $\left(x,0\right)$ is an $x$ -intercept of $f$ .

Thus, $x$ is a zero of $f$ if and only if $\left(x,0\right)$ is an $x$ -intercept of $f$ .

The function $f\left(x\right)=x^4+3x^3-9x^2+2x+3$is graphed along with the $x$

intercepts:

  

The function $f\left(x\right)=x^4+3x^3-9x^2+2x+3$ is coloured in red and the intercepts are coloured in grey.

Estimate the zeroes of $f\left(x\right)=x^4+3x^3-9x^2+2x+3$from the graph:

It is clear from the graph that it has only four $x$ -intercepts:

  $(-4.9,0)$ , $(-0.45,0)$ , $(1,0)$ and $(1.35,0)$ .

Thus, the zeroes of the function $f\left(x\right)=x^4+3x^3-9x^2+2x+3$ are approximately:

  $x=-4.9$ , $x=-0.45$ , $x=1$ and $x=1.35$ .

Conclusion:

The zeroes of the function $f\left(x\right)=x^4+3x^3-9x^2+2x+3$ are approximately:

  $x=-4.9$ , $x=-0.45$ , $x=1$ and $x=1.35$ .