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 \[f(x)=x^3-3\cdot x\]
The zeros of #f(x)# are #x=0 \lor x=\sqrt{3} \lor x=-\sqrt{3}#.
The derivative is #f'(x)=##3\cdot x^2-3#.
The stationary points of #f(x)# are #x=-1 \lor x=1#.
The corresponding extreme values are #f\left(-1\right)=# #2# and #f\left(1\right)=# #-2#.
The sign diagram of #f'(x)# looks like this:
The increasing intervals of #f(x)# are #\ivoo{-\infty}{-1}# and #\ivoo{1}{\infty}#; the decreasing interval is #\ivoo{-1}{1}#.
The graph of #f(x)# looks like this:
As for extreme points: at #x=-1# the function #f(x)# has a local maximum #2# and at #x=1# a local minimum #-2#. These local extrema are not global.
First, we calculate the zeros of #f(x)#:
\[\begin{array}{rcll}
f(x) =0&&&\phantom{xx}\color{blue}{\text{the equation we have to solve }} \\
x^3-3\cdot x =0&&&\phantom{xx}\color{blue}{\text{function rule entered }} \\
x \cdot \left(x^2-3\right) =0&&&\phantom{xx}\color{blue}{\text{left hand side factored}} \\
x=0 \lor x^2-3=0&&&\phantom{xx}\color{blue}{A\cdot B^2=0\Leftrightarrow A=0\lor B=0} \\
x=0 \lor x=\sqrt{3} \lor x=-\sqrt{3}&&&\phantom{xx}\color{blue}{A^2=a\Leftrightarrow A=\sqrt{a} \lor A=-\sqrt{a}} \\
\end{array}\]
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)-3 \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:
\[\begin{array} {rcl} f'(x)&=&\frac{\dd}{\dd x}\left( x^3-3\cdot x\right)\\
&=&\frac{\dd}{\dd x}\left(x^3\right)-3 \cdot\frac{\dd}{\dd x}\left (x\right)\\
&&\phantom{xx}\color{blue}{\text{sum rule}}\\
&=&
3 \cdot x^{3-1} - 3 \cdot x^{1-1} \\
&&\phantom{xx}\color{blue}{\text{power rule}}\\
&=&3\cdot x^2-3 \end{array}\]
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^2-3# we can find the stationary points the following way:
\[\begin{array}{rl}
3\cdot x^2-3=0&\phantom{xxx}\color{blue}{\text{equation entered}}\\
3\cdot x^2=3&\phantom{xxx}\color{blue}{-3 \text{ moved to the other side}}\\
x^2=1&\phantom{xxx}\color{blue}{\text{divided by 3}}\\
x=-1 \lor x=1&\phantom{xxx}\color{blue}{A^2=a\Leftrightarrow A=\sqrt{a} \lor A=-\sqrt{a}}\end{array}
\]
Next we find the corresponding values of #f(x)# with the stationary points. We find the value of #f# at #x=-1# by entering #x=-1# in #f(x)#:
\[f\left(-1\right)=\left(-1\right)^3-3 \cdot \left(-1\right)=2\tiny.\]
We determine the value of #f# at #x=1# in the same manner:
\[f\left(1\right)=\left(1\right)^3-3 \cdot \left(1\right)=-2\tiny.\]
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 #-1# and the biggest one on the right, which is #1#.
If you take a look at the sign diagram of #f#, you see it has plus signs at the left up to #-1# and from #1# till the end. Hence, on the intervals #\ivoo{-\infty}{-1}# and #\ivoo{1}{\infty}#, the function #f# is increasing. On the part between #-1# and #1# the diagram has minuses, hence, on the interval #\ivoo{-1}{1}#, the function #f# is decreasing. This means that #\ivoo{-\infty}{-1}# and #\ivoo{1}{\infty}# are the increasing intervals of #f(x)#; the decreasing interval is #\ivoo{-1}{1}#.
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=-1# the function #f(x)# has a local maximum #2# and at #x=1# a local minimum #-2#. These local extrema are not global.
The derivative is #f'(x)=##3\cdot x^2-3#.
The stationary points of #f(x)# are #x=-1 \lor x=1#.
The corresponding extreme values are #f\left(-1\right)=# #2# and #f\left(1\right)=# #-2#.
The sign diagram of #f'(x)# looks like this:
The increasing intervals of #f(x)# are #\ivoo{-\infty}{-1}# and #\ivoo{1}{\infty}#; the decreasing interval is #\ivoo{-1}{1}#.
The graph of #f(x)# looks like this:
As for extreme points: at #x=-1# the function #f(x)# has a local maximum #2# and at #x=1# a local minimum #-2#. These local extrema are not global.
First, we calculate the zeros of #f(x)#:
\[\begin{array}{rcll}
f(x) =0&&&\phantom{xx}\color{blue}{\text{the equation we have to solve }} \\
x^3-3\cdot x =0&&&\phantom{xx}\color{blue}{\text{function rule entered }} \\
x \cdot \left(x^2-3\right) =0&&&\phantom{xx}\color{blue}{\text{left hand side factored}} \\
x=0 \lor x^2-3=0&&&\phantom{xx}\color{blue}{A\cdot B^2=0\Leftrightarrow A=0\lor B=0} \\
x=0 \lor x=\sqrt{3} \lor x=-\sqrt{3}&&&\phantom{xx}\color{blue}{A^2=a\Leftrightarrow A=\sqrt{a} \lor A=-\sqrt{a}} \\
\end{array}\]
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)-3 \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:
\[\begin{array} {rcl} f'(x)&=&\frac{\dd}{\dd x}\left( x^3-3\cdot x\right)\\
&=&\frac{\dd}{\dd x}\left(x^3\right)-3 \cdot\frac{\dd}{\dd x}\left (x\right)\\
&&\phantom{xx}\color{blue}{\text{sum rule}}\\
&=&
3 \cdot x^{3-1} - 3 \cdot x^{1-1} \\
&&\phantom{xx}\color{blue}{\text{power rule}}\\
&=&3\cdot x^2-3 \end{array}\]
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^2-3# we can find the stationary points the following way:
\[\begin{array}{rl}
3\cdot x^2-3=0&\phantom{xxx}\color{blue}{\text{equation entered}}\\
3\cdot x^2=3&\phantom{xxx}\color{blue}{-3 \text{ moved to the other side}}\\
x^2=1&\phantom{xxx}\color{blue}{\text{divided by 3}}\\
x=-1 \lor x=1&\phantom{xxx}\color{blue}{A^2=a\Leftrightarrow A=\sqrt{a} \lor A=-\sqrt{a}}\end{array}
\]
Next we find the corresponding values of #f(x)# with the stationary points. We find the value of #f# at #x=-1# by entering #x=-1# in #f(x)#:
\[f\left(-1\right)=\left(-1\right)^3-3 \cdot \left(-1\right)=2\tiny.\]
We determine the value of #f# at #x=1# in the same manner:
\[f\left(1\right)=\left(1\right)^3-3 \cdot \left(1\right)=-2\tiny.\]
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 #-1# and the biggest one on the right, which is #1#.
- Next you enter an #x \lt -1# in #f'(x)#. For example #x=-10#. Then you get #f'(-10)=3 \cdot (-10)^2 -3 =297#. Because #f'(x) \gt 0#, #f(x)# increases on the interval #x \lt -1# and you write down ++++.
- Next you enter a #-1 \lt x \lt 1# in #f'(x)#. For example #x=0#. Then you get #f'(0)=3 \cdot (0)^2 -3 =-3#. Because #f'(x) \lt 0#, #f(x)# decreases on the interval #-1 \lt x \lt 1# and you write down ----.
- Next you enter an #x \gt 1# in #f'(x)#. For example #x=10#. Then you get #f'(10)=3 \cdot (10)^2 - 3 =297#. Because #f'(x) \gt 0#, #f(x)# increases on the interval #x \gt 1# 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 #-1# and from #1# till the end. Hence, on the intervals #\ivoo{-\infty}{-1}# and #\ivoo{1}{\infty}#, the function #f# is increasing. On the part between #-1# and #1# the diagram has minuses, hence, on the interval #\ivoo{-1}{1}#, the function #f# is decreasing. This means that #\ivoo{-\infty}{-1}# and #\ivoo{1}{\infty}# are the increasing intervals of #f(x)#; the decreasing interval is #\ivoo{-1}{1}#.
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=-1# the function #f(x)# has a local maximum #2# and at #x=1# a local minimum #-2#. These local extrema are not global.
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