Question

Analyze each polynomial function using Steps 1 through 8 on page 190 in Section 4.1
$f\left(x\right)=4x^5+12x^4-x-3$

Short answer

For the given polynomial function, we get:

x-intercept$=-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},-3$

y-intercept$=-3$

Step-by-step solution

Step 1. End behavior of the function

We have a degree of five from $f\left(x\right)=4x^5+12x^4-x-3$.

Thus, the graph of function has an end behavior similar to $y=4x^5$.

Step 2. Determine the intercept of the function

Determine the intercept of function to get the x- intercept.

$\begin{array}{r}4x^5+12x^4-x-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ 4x^4(x+3)-(x+3)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \left(4x^4-1\right)(x+3)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ 2x^2+1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0;\begin{array}{rlrlrl}\left(2x^2+1\right)\left(2x^2-1\right)(x+3)& =0& & & & \\ 2x^2-1& =0;x+3& =0& & & \\ 2x^2& =-1;& 2x^2& =1;& x& =-3\\ \sqrt{x^2}& =-\sqrt{\frac{1}{2};}& \sqrt{x^2}& =\sqrt{\frac{1}{2}}& & \\ x& =-\frac{\sqrt{2}}{2};& x& =\frac{\sqrt{2}}{2}& & \end{array}\end{array}$

Considering $f\left(0\right)=-3$we have y-intercept$(0,-3)$

Step 3. Graph the function

Step 4. Maximum and minimum turning point

From the graph above, we can see that the maximum and minimum turning point of the function is:

$(-2.404,79.028)$and$(0.266,-3.201)$

Step 5. Graph of function with obtained result

Step 6. Domain and range

Therefore, the domain and range of the function is the set of all real numbers.

Step 7. Interval at which the graph is increasing or decreasing

By observing the graph above, we say that the function at the intervals of $(-\infty ,2.404)$and $(0.266,\infty )$is increasing.

At intervals of $(-2.404,0.266)$decreasing.