Analysis of functions

Applications of differentiation: Analysis of functions

Analysis of functions

In the previous paragraphs, we have seen how you can determine if a function increases or decreases on the basis of the derivative, that extremes always are a stationary point, in which 0 the derivative is 0.

With the help of these properties a function $f(x)$ can be analyzed. You do this by following these seven steps.

Determine all zeros of $f(x)$ .
Calculate the derivative function $f'(x)$ .
Determine the stationary points of $f(x)$ (ie the zeros of $f'(x)$ ).
Calculate the function value $f(x)$ in each stationary point $x$ .
Create a sign diagram of $f'(x)$ . This is a diagram showing for values of $x$ whether the function $f(x)$ increases (which you would indicate with ++++) or decreases (which you would indicate with ----). The stationary points in the diagram are indicated with a 0, because in these the derivative equal to 0.
Make an indication of the intervals where the function is increasing / decreasing.
Make an accurate drawing of the graph of the function $f(x)$ .

Analyse the function $f(x)=x^3-12\cdot x$

The zeros of $f(x)$ are $x=0 \lor x=2\cdot \sqrt{3} \lor x=-2\cdot \sqrt{3}$ .
The derivative is $f'(x)=$ $3\cdot x^2-12$ .
The stationary points of $f(x)$ are $x=-2 \lor x=2$ .
The corresponding extreme values are $f\left(-2\right)=$ $16$ and $f\left(2\right)=$ $-16$ .
The sign diagram of $f'(x)$ looks like this:

The increasing intervals of $f(x)$ are $\ivoo{-\infty}{-2}$ and $\ivoo{2}{\infty}$ ; the decreasing interval is $\ivoo{-2}{2}$ .
The graph of $f(x)$ looks like this:

As for extreme points: at $x=-2$ the function $f(x)$ has a local maximum $16$ and at $x=2$ a local minimum $-16$ . These local extrema are not global.

First, we calculate the zeros of $f(x)$ :
$\begin{array}{rcll} f(x) =0&&&\phantom{xx}\color{blue}{\text{the equation we have to solve }} \\ x^3-12\cdot x =0&&&\phantom{xx}\color{blue}{\text{function rule entered }} \\ x \cdot \left(x^2-12\right) =0&&&\phantom{xx}\color{blue}{\text{left hand side factored}} \\ x=0 \lor x^2-12=0&&&\phantom{xx}\color{blue}{A\cdot B^2=0\Leftrightarrow A=0\lor B=0} \\ x=0 \lor x=2\cdot \sqrt{3} \lor x=-2\cdot \sqrt{3}&&&\phantom{xx}\color{blue}{A^2=a\Leftrightarrow A=\sqrt{a} \lor A=-\sqrt{a}} \\ \end{array}$

Next, we calculate the derivative of $f(x)$ . To this end we use the extended sum rule. It says that $f'(x)=\frac{\dd}{\dd x}\left(x^3\right)-12 \cdot\frac{\dd}{\dd x}\left (x\right)$ .
With help of the polynomial rule for differentiation, which says that $\frac{\dd}{\dd x}\left(x^n\right)=n \cdot x^{n-1}$ we now have:
$\begin{array} {rcl} f'(x)&=&\frac{\dd}{\dd x}\left( x^3-12\cdot x\right)\\ &=&\frac{\dd}{\dd x}\left(x^3\right)-12 \cdot\frac{\dd}{\dd x}\left (x\right)\\ &&\phantom{xx}\color{blue}{\text{sum rule}}\\ &=& 3 \cdot x^{3-1} - 12 \cdot x^{1-1} \\ &&\phantom{xx}\color{blue}{\text{power rule}}\\ &=&3\cdot x^2-12 \end{array}$

Now we calculate the stationary points of $f(x)$ . Stationary points are the points for which we have $f'(x)=0$ . Since $f'(x)=3\cdot x^2-12$ we can find the stationary points the following way:
$\begin{array}{rl} 3\cdot x^2-12=0&\phantom{xxx}\color{blue}{\text{equation entered}}\\ 3\cdot x^2=12&\phantom{xxx}\color{blue}{-12 \text{ moved to the other side}}\\ x^2=4&\phantom{xxx}\color{blue}{\text{divided by 3}}\\ x=-2 \lor x=2&\phantom{xxx}\color{blue}{A^2=a\Leftrightarrow A=\sqrt{a} \lor A=-\sqrt{a}}\end{array}$

Next we find the corresponding values of $f(x)$ with the stationary points. We find the value of $f$ at $x=-2$ by entering $x=-2$ in $f(x)$ :
$f\left(-2\right)=\left(-2\right)^3-12 \cdot \left(-2\right)=16\tiny.$
We determine the value of $f$ at $x=2$ in the same manner:
$f\left(2\right)=\left(2\right)^3-12 \cdot \left(2\right)=-16\tiny.$

Now we can make a sign diagram for $f(x)$ . In the stationary points of $f$ , we have $f'(x)=0$ . Place the smallest stationary point on the left hand side, which is $-2$ and the biggest one on the right, which is $2$ .

Next you enter an $x \lt -2$ in $f'(x)$ . For example $x=-10$ . Then you get $f'(-10)=3 \cdot (-10)^2 -12 =288$ . Because $f'(x) \gt 0$ , $f(x)$ increases on the interval $x \lt -2$ and you write down ++++.
Next you enter a $-2 \lt x \lt 2$ in $f'(x)$ . For example $x=0$ . Then you get $f'(0)=3 \cdot (0)^2 -12 =-12$ . Because $f'(x) \lt 0$ , $f(x)$ decreases on the interval $-2 \lt x \lt 2$ and you write down ----.
Next you enter an $x \gt 2$ in $f'(x)$ . For example $x=10$ . Then you get $f'(10)=3 \cdot (10)^2 - 12 =288$ . Because $f'(x) \gt 0$ , $f(x)$ increases on the interval $x \gt 2$ and you write down ++++.

This leads to the following sign diagram:

If you take a look at the sign diagram of $f$ , you see it has plus signs at the left up to $-2$ and from $2$ till the end. Hence, on the intervals $\ivoo{-\infty}{-2}$ and $\ivoo{2}{\infty}$ , the function $f$ is increasing. On the part between $-2$ and $2$ the diagram has minuses, hence, on the interval $\ivoo{-2}{2}$ , the function $f$ is decreasing. This means that $\ivoo{-\infty}{-2}$ and $\ivoo{2}{\infty}$ are the increasing intervals of $f(x)$ ; the decreasing interval is $\ivoo{-2}{2}$ .

With this information, the graph of $f(x)$ can be drawn. It is shown near the top of this solution. In addition, you can now identify the extreme points: at $x=-2$ the function $f(x)$ has a local maximum $16$ and at $x=2$ a local minimum $-16$ . These local extrema are not global.

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