- Problem
- Detailed Solution
- Summary Solution

### Extrema

Determine minima and/or maxima of the function

\(k(t)=12-12t+t^3\).

### Preliminary remarks

You are asked to localize extrema of the function \(k(t)=12-12t+t^3\), and to find out which one is a minimum and which one is a maximum.

To do this, you need the following competences:

- Be able to determine „candidate t-values“ for an extremum (see script block III learning sequence 2, theorem III.2.6.3)
- Make use of the 2
^{nd}order derivative to find out whether there actually is an extremum, and to know whether it is a minimum or a maximum (see script block III learning sequence 2, theorem III.2.6.4).

### Formal approach

*(i) Determine "candidate
t-values" for an extremum of \(k(t)\)
*

To find such candidate values, determine the 1^{st} order
derivative of k(t) and look for the roots of k'. The reason for
this is that the tangent to a graph at an extremum of that graph
must be horizontal, and a horizontal straight line has a slope of
zero.

\(k'(t)=-12+3t^2=0 \quad \Rightarrow \quad 3t^2=12 \quad \Rightarrow \quad t^2=4\)

\( \Rightarrow \quad {t_1}=+2\) und \({t_2}=-2\)

Obviously, there are two different places \({t_1}=+2\) and \({t_2}=-2\) on the horizontal axis where k might have an extremum.

*(ii) Make use of the 2nd order derivative ( maximum
or minimum?)
*

\(k(t)\) has a **maximum** at a candidate t-value if,
for that value, the 2^{nd} order derivative **\(k''(t)\)
is negative**. This last condition means that, when
travelling along the graph of k, in the moment of passing by the
candidate value, we are in a right bent.

Likewise, \(k(t)\)) has a **minimum** at a candidate
t-value if, for that value, the 2^{nd} order derivative **\(k''(t)\)
is positive**. This last condition means that, when
travelling along the graph of k, in the moment of passing by the
candidate value, we are in a left bent.

By the way, the « candidate values » are called « critical values » or « stationary values ».

We now know how to proceed:

The 2^{nd} order derivative of \(k(t)\) is \(\;k''(t)=6t\)

We evaluate this function for the critical (i.e. stationary) values:

- For \({t_1}=+2\), we obtain \(k''({t_1})=k''(2)=6 \cdot 2=12>0\).

Thus, we have a positive value (+ 12) of k''. (Note that the absolute amount of the 2nd order derivative is unimportant, we need only the sign). Thus, we can conclude that at \({t_1}=+2\) , k is concave up, and so k has a local minimum at that place.

- For the second critical value \({t_2}=-2\), we have \(k''({t_2})=k''(-2)=6 \cdot (-2)=-12 < 0 \).

Again, only the sign of this result is important, and since it is negative, we conclude that at \({t_2}=-2\), k is concave down, so k has a local maximum at that place.

After this work has been done, we can have a look at the diagram to check our results.

To sum up our approach, we needed the roots of \(k'(t)\) to find
critical values for an extremum of k, and then we determined the
sign of \(k''(t)\) in these critical places to be sure there
actually is an extremum, and to tell whether it is a minimum or a
maximum.

A visualization of these relations is provided here.

\(k'(t)=-12+3t^2=0 \quad \Rightarrow \quad {t_1}=+2 \quad {t_2}=-2\)

\(k''(t)=6t\)

\(k''({t_1})=k''(2)=6 \cdot 2=12 > 0 \quad \Rightarrow \quad \) k has a local minimum at \({t_1}=+2\).

\(k''({t_2})=k''(-2)=6 \cdot (-2)=-12 < 0 \quad \Rightarrow \quad \) k has a local maximum at \({t_2}=-2\).