- 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 2^{nd}
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 2^{nd} 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 2^{nd} 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).