# Curve Sketching with Derivatives?

1. Mar 24, 2008

### Delber

1. The problem statement, all variables and given/known data
I am given the following function:

$$f(x)=\frac{2x^2}{x-1}$$

The question asks to find the critical points and classify each using the second derivative.

2. Relevant equations

3. The attempt at a solution
I got the derivative of f(x) to be:
$$f'(x)=\frac{2x^2-4x}{(x-1)^2}$$
So the critical points are when x=0, x=1, and x=2.

Here's where I'm stuck. I don't know what it means when it asks to classify each using the second derivative.
If it helps, the second derivative is:

$$f''(x)=\frac{-8x^2+12x-4}{(x-1)^4}$$

Last edited: Mar 24, 2008
2. Mar 24, 2008

### jhicks

What it means is that you can use the second derivative to determine if there is either a local minimum or local maximum or asymptote. For example, if f'(x)=0 and f''(x) is positive, you know that there is a local minimum.

3. Mar 24, 2008

### BrendanH

Second derivative determines concavity or convexity of a function. A function whose second derivative is everywhere positive is increasing at an increasing rate. This makes it convex downward, or just concave. On the other hand, a negative second derivative indicates a convex upward. When the 2nd derivative is zero, this is an inflection point.