• 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 2nd 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 1st 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 2nd 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 2nd 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 2nd 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 ${t_2}=-2$.