Thread: Instantaneous rate of change: determining where it is positive, negative, and zero

1. Instantaneous rate of change: determining where it is positive, negative, and zero

Hi,

I hope someone can help. I'm trying to find the x-coordinate for the turning point (circled in red) within my drawing:

This drawing is based on the equation f(x) = x^3-2x^2+x

I need to know this x-coordinate in order for me to accurately state the intervals where the rate of change is positive and negative. Since turning points indicate an opposite change in slope, I therefore need to know what the x-coordinate of this turning point is.

I know that it's a cubic function that has a positive leading coefficient, therefore having an end behaviour that extends from quadrant 1 to quadrant 3. I also know that the roots are x = 0 and x = 1 (with a multiplicity of 2). I also know that I could use graphing technology (e.g. Demos), but I want to know if there's a simple way to find the x-coordinate for this turning point algebraically.

Sincerely,
Olivia

2. Re: Instantaneous rate of change: determining where it is positive, negative, and zer

If a function has turning points, they occur where $f'(x)=0$.
This pre-supposes you know a little bit of calculus.

If you don't, then use a calculator to determine the local max of $f(x)$ at that point.

3. Re: Instantaneous rate of change: determining where it is positive, negative, and zer

Sounds good. Thanks !