## Curve Sketching (General Scheme)

1. First Attempt. We have some complicated function, and we want to graph it…but don’t have a calculator, computer, cell phone, etc. What do we do?

Lets fix our notation now. We are trying to sketch the curve of $f(x)$.

Well, we know that if $f'(x)>0$ on some interval $(a,b)$, then it is “increasing”. (Meaning: if $b>x_{1}>x_{0}>a$, then $f(x_{1})>f(x_{0})$) At worst, we could just draw a straight line from $(a,f(a))$ to $(b,f(b))$.

Similar reasoning suggests we could do likewise if $f'(x)<0$ on some interval.

How do we find these intervals? We just find the critical points $x_{j}$ satisfying $f'(x_{j})=0$.

This gives us a very rough sketch. So to reiterate, the algorithm is:

1. Find the critical points $x_{j}\in\mathbb{R}$.
2. Find the sign of $f'(x)$ on the intervals $(x_{j}, x_{j+1})$.
3. Find the values of the points $f(x_{j})$ for each $x_{j}$. (Or just guess some appropriate value, if that's easier.)
4. Sketch the curve as a collection of straight lines.

The question: can we do better?

2. Second Derivative Test. We can tell something about the nature of the critical points by examining the second derivatives there.

Let $x_{j}$ describe a critical point of $f$.
(a) If $f''(x_{j})>0$, then $x_{j}$ is a local maximum.
(b) If $f''(x_{j})<0$, then $x_{j}$ is a local minimum.
(c) If $f''(x_{j})=0$, then we don't know đŸ˜•

This idea is good, and a first step, but needs gives us little information.

3. Second Derivatives. What we can do is find the inflection points, i.e., the points where $f''(c)=0$. Then, using our solution from exercise 5 of implicit differentation, we just paste together portions of a circle, and that’s our curve.

The four quadrants give us the segments when the circle is increasing or decreasing, and concave up or concave down. The four situations are drawn with the three tangent lines drawn in light red:

 Quadrant II: Concave down and increasing. Quadrant I: Concave down and decreasing.

We should note that the concave down segments looks like a bowl turned upside down. The concave up segments looks like a right-side up bowl.

 Quadrant III: Concave up and decreasing. Quadrant IV: Concave up and increasing.

We now glue these together according to information we obtain using the function’s first and second derivatives. We’ll consider examples in the next post.