>Applications of the Derivative>How to Analyze Polynomial Functions

How to Analyze Polynomial Functions

Let’s look at an example of analysis of a polynomial function. The method is as follows:

Rule

Analyzing Polynomial Functions

1.: Find the zeros.
2.: Find the stationary points.
3.: Find the inflection points.

Example 1

Analyze the function $f (x) = x^{3} + 6 x^{2} + 8 x$

1.

Find the zeros of the function by setting

f (x) = 0

. Since this is a polynomial function of degree 3, you cannot use the quadratic formula to solve it directly. However, as

f (x)

does not have a constant term, we can factorize

x

out of the rest of the function:

\begin{array}{l} f (x) & = x^{3} + 6 x^{2} + 8 x \\ = x (x^{2} + 6 x + 8) \end{array}

\begin{array}{l} f (x) = x^{3} + 6 x^{2} + 8 x = x (x^{2} + 6 x + 8) \end{array}

This means that

x = 0

is a zero. You can then use the quadratic formula formula on

x^{2} + 6 x + 8

to find the other zeros:

\begin{array}{l} x & = \frac{- 6 \pm \sqrt{(6^{2} - 4 \cdot 1 \cdot 8}}{2} \\ = \frac{- 6 \pm \sqrt{4}}{2} \\ = \frac{- 6 \pm 2}{2} \end{array}

Thus, $x = - 4$ or $x = - 2$ . The zeros of $f (x)$ are thus $x = - 4$ , $x = - 2$ and $x = 0$ .

2.

Find the maxima and minima by setting

f^{'} (x) = 0

. Find the derivative of

f (x) = x^{3} + 6 x^{2} + 8 x

f^{'} (x) = 3 x^{2} + 12 x + 8

Then use the quadratic formula to find the maxima and minima of $f (x)$ : $\begin{array}{l} x & = \frac{- 12 \pm \sqrt{(1 2^{2} - 4 \cdot 1 \cdot 8}}{2} \\ = \frac{- 12 \pm \sqrt{112}}{2} \\ \approx \frac{- 6 \pm 10.58}{2} \end{array}$

Thus, $x \approx - 3.15$ or $x \approx - 0.85$ . To find the points, you need to find their corresponding $y$ -values. You find these by putting the $x$ -values you found back into the main function $f (x)$ : $\begin{array}{l} y & = f (- 3.15) \approx 3.08 \\ y & = f (- 0.85) \approx - 3.08 \end{array}$

You now need to determine which point is a maximum and which is a minimum. You do that by drawing a sign chart. Notice that the derivative can be factorized as:

f^{'} (x) = (x + 3.15) (x + 0.85)

Sign chart of a polynomial function

From this, you can see that the maximum is situated at

(- 3.15, 3.08)

and the minimum is situated at

(- 0.85, - 3.08)

3.

Find the inflection points by setting

f^{″} (x) = 0

. First, you find the second derivative of

f (x)

by differentiating

f^{'} (x) = 3 x^{2} + 12 x + 8

f^{″} (x) = 6 x + 12 .

Let $f^{″} (x) = 0$ and solve the equation:

\begin{array}{l} 6 x + 12 & = 0 \\ 6 x & = - 12 \\ x & = - 2 \end{array}

Enter this $x$ -value into the original function $f (x)$ to find the $y$ -coordinate for the inflection point:

\begin{array}{l} f (- 2) & = {(- 2)}^{3} + 6 \cdot {(- 2)}^{2} + 8 \cdot (- 2) \\ = - 8 + 24 - 16 \\ = 0 \end{array}

\begin{array}{l} f (- 2) & = {(- 2)}^{3} + 6 \cdot {(- 2)}^{2} + 8 \cdot (- 2) \\ = - 8 + 24 - 16 \\ = 0 \end{array}

The inflection point is thus

(- 2, 0)

. By making a sign chart for the second derivative, you can see where the graph of

f (x)

is concave and where it is convex. Notice that

f^{″} (x)

can be factorized as

f^{″} (x) = 6 (x + 2)

Sign chart of f”(x) = 6x+12.

What Are Inflection Points of a Function?

How to Analyze Logarithmic Functions

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