• 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 $${t_2}=-2$$.