• Problem
• Detailed Solution
• Summary Solution

### Curvature

Name intervals on which the following function is concave up or down:

$K(x)=x^3-2x^2+60x+100$

### Preliminary remarks

The problem consists of analyzing the curvature of a function. To solve it, you need to know how curvature, i.e. the quality of being concave up or concave down, is characterized by the 2nd order derivative of a function (see theorem III.2.5.2 in block III learning sequence 2, of our script).

### Formal Approach

(i) Determine sections where this function is concave up

To find an answer, we need the 2nd order derivative of K(x). Then, we determine values x for which K''(x) is positive.

$K'(x)=3x^2-4x+60 \quad \Rightarrow \quad K''(x)=6x-4$

$K''(x)=6x-4 > 0 \quad \Rightarrow \quad x > 2/3$

Our conclusion: for x > 2/3, K is concave up.
(Note that the sign of K'' characterizes the curvature of K, not the curvature of K' or of K'' itself!).

(ii) Determine sections where this function is concave up

Here again, we need the 2nd order derivative, which we now know is 6x−4. This time, we look for sections where K''(x) is negative:

$K'(x)=3x^2-4x+60 \quad \Rightarrow \quad K''(x)=6x-4$.

$K''(x)=6x-4 < 0 \quad \Rightarrow \quad x < 2/3$

Thus, we conclude that for x < 2/3, K is concave down.
(Note again that K’’ characterizes the curvature of K).

You can check your findings by sketching the graph of K, or by looking at the diagram.

Please keep in mind that our problem could be solved by analyzing values of K''. To visualize the relation between the curvature of K and the sign of its 2nd order derivative, please have a look at this second diagram.

$K'(x)=3x^2-4x+60$

$K''(x)=6x-4$

Concave up:

$K''(x)=6x-4 > 0 \quad \Rightarrow \quad x > 2/3 \quad \Rightarrow \quad$ K(x) is concave up for $x>2/3$ (left hand bent).

Concave down:

$K''(x)=6x-4 < 0 \quad \Rightarrow \quad x < 2/3 \quad \Rightarrow \quad$ K(x) ist concave down for $x < 2/3$ (right hand bent).