### 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 \[f(x)=x^3-48\cdot x\]

The zeros of #f(x)# are #x=0 \lor x=4\cdot \sqrt{3} \lor x=-4\cdot \sqrt{3}#.

The derivative is #f'(x)=##3\cdot x^2-48#.

The stationary points of #f(x)# are #x=-4 \lor x=4#.

The corresponding extreme values are #f\left(-4\right)=# #128# and #f\left(4\right)=# #-128#.

The sign diagram of #f'(x)# looks like this:

The increasing intervals of #f(x)# are #\ivoo{-\infty}{-4}# and #\ivoo{4}{\infty}#; the decreasing interval is #\ivoo{-4}{4}#.

The graph of #f(x)# looks like this:

As for extreme points: at #x=-4# the function #f(x)# has a local maximum #128# and at #x=4# a local minimum #-128#. 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^3-48\cdot x =0&&&\phantom{xx}\color{blue}{\text{function rule entered }} \\

x \cdot \left(x^2-48\right) =0&&&\phantom{xx}\color{blue}{\text{left hand side factored}} \\

x=0 \lor x^2-48=0&&&\phantom{xx}\color{blue}{A\cdot B^2=0\Leftrightarrow A=0\lor B=0} \\

x=0 \lor x=4\cdot \sqrt{3} \lor x=-4\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

With help of the

\[\begin{array} {rcl} f'(x)&=&\frac{\dd}{\dd x}\left( x^3-48\cdot x\right)\\

&=&\frac{\dd}{\dd x}\left(x^3\right)-48 \cdot\frac{\dd}{\dd x}\left (x\right)\\

&&\phantom{xx}\color{blue}{\text{sum rule}}\\

&=&

3 \cdot x^{3-1} - 48 \cdot x^{1-1} \\

&&\phantom{xx}\color{blue}{\text{power rule}}\\

&=&3\cdot x^2-48 \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^2-48# we can find the stationary points the following way:

\[\begin{array}{rl}

3\cdot x^2-48=0&\phantom{xxx}\color{blue}{\text{equation entered}}\\

3\cdot x^2=48&\phantom{xxx}\color{blue}{-48 \text{ moved to the other side}}\\

x^2=16&\phantom{xxx}\color{blue}{\text{divided by 3}}\\

x=-4 \lor x=4&\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=-4# by entering #x=-4# in #f(x)#:

\[f\left(-4\right)=\left(-4\right)^3-48 \cdot \left(-4\right)=128\tiny.\]

We determine the value of #f# at #x=4# in the same manner:

\[f\left(4\right)=\left(4\right)^3-48 \cdot \left(4\right)=-128\tiny.\]

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 #-4# and the biggest one on the right, which is #4#.

If you take a look at the sign diagram of #f#, you see it has plus signs at the left up to #-4# and from #4# till the end. Hence, on the intervals #\ivoo{-\infty}{-4}# and #\ivoo{4}{\infty}#, the function #f# is increasing. On the part between #-4# and #4# the diagram has minuses, hence, on the interval #\ivoo{-4}{4}#, the function #f# is decreasing. This means that #\ivoo{-\infty}{-4}# and #\ivoo{4}{\infty}# are the increasing intervals of #f(x)#; the decreasing interval is #\ivoo{-4}{4}#.

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=-4# the function #f(x)# has a local maximum #128# and at #x=4# a local minimum #-128#. These local extrema are not global.

The derivative is #f'(x)=##3\cdot x^2-48#.

The stationary points of #f(x)# are #x=-4 \lor x=4#.

The corresponding extreme values are #f\left(-4\right)=# #128# and #f\left(4\right)=# #-128#.

The sign diagram of #f'(x)# looks like this:

The increasing intervals of #f(x)# are #\ivoo{-\infty}{-4}# and #\ivoo{4}{\infty}#; the decreasing interval is #\ivoo{-4}{4}#.

The graph of #f(x)# looks like this:

As for extreme points: at #x=-4# the function #f(x)# has a local maximum #128# and at #x=4# a local minimum #-128#. 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^3-48\cdot x =0&&&\phantom{xx}\color{blue}{\text{function rule entered }} \\

x \cdot \left(x^2-48\right) =0&&&\phantom{xx}\color{blue}{\text{left hand side factored}} \\

x=0 \lor x^2-48=0&&&\phantom{xx}\color{blue}{A\cdot B^2=0\Leftrightarrow A=0\lor B=0} \\

x=0 \lor x=4\cdot \sqrt{3} \lor x=-4\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)-48 \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^3-48\cdot x\right)\\

&=&\frac{\dd}{\dd x}\left(x^3\right)-48 \cdot\frac{\dd}{\dd x}\left (x\right)\\

&&\phantom{xx}\color{blue}{\text{sum rule}}\\

&=&

3 \cdot x^{3-1} - 48 \cdot x^{1-1} \\

&&\phantom{xx}\color{blue}{\text{power rule}}\\

&=&3\cdot x^2-48 \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^2-48# we can find the stationary points the following way:

\[\begin{array}{rl}

3\cdot x^2-48=0&\phantom{xxx}\color{blue}{\text{equation entered}}\\

3\cdot x^2=48&\phantom{xxx}\color{blue}{-48 \text{ moved to the other side}}\\

x^2=16&\phantom{xxx}\color{blue}{\text{divided by 3}}\\

x=-4 \lor x=4&\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=-4# by entering #x=-4# in #f(x)#:

\[f\left(-4\right)=\left(-4\right)^3-48 \cdot \left(-4\right)=128\tiny.\]

We determine the value of #f# at #x=4# in the same manner:

\[f\left(4\right)=\left(4\right)^3-48 \cdot \left(4\right)=-128\tiny.\]

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 #-4# and the biggest one on the right, which is #4#.

- Next you enter an #x \lt -4# in #f'(x)#. For example #x=-10#. Then you get #f'(-10)=3 \cdot (-10)^2 -48 =252#. Because #f'(x) \gt 0#, #f(x)# increases on the interval #x \lt -4# and you write down ++++.
- Next you enter a #-4 \lt x \lt 4# in #f'(x)#. For example #x=0#. Then you get #f'(0)=3 \cdot (0)^2 -48 =-48#. Because #f'(x) \lt 0#, #f(x)# decreases on the interval #-4 \lt x \lt 4# and you write down ----.
- Next you enter an #x \gt 4# in #f'(x)#. For example #x=10#. Then you get #f'(10)=3 \cdot (10)^2 - 48 =252#. Because #f'(x) \gt 0#, #f(x)# increases on the interval #x \gt 4# and you write down ++++.

If you take a look at the sign diagram of #f#, you see it has plus signs at the left up to #-4# and from #4# till the end. Hence, on the intervals #\ivoo{-\infty}{-4}# and #\ivoo{4}{\infty}#, the function #f# is increasing. On the part between #-4# and #4# the diagram has minuses, hence, on the interval #\ivoo{-4}{4}#, the function #f# is decreasing. This means that #\ivoo{-\infty}{-4}# and #\ivoo{4}{\infty}# are the increasing intervals of #f(x)#; the decreasing interval is #\ivoo{-4}{4}#.

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=-4# the function #f(x)# has a local maximum #128# and at #x=4# a local minimum #-128#. These local extrema are not global.

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