I have to find the relative minimum and relative maximum of g'(x)=(x+1)(x-1)^2(x-2). What I've done so far is constructed a sign chart using the critical numbers, but I'm confused. To the left of -1 it's +, to the right of -1 it's -, to the left of 1 it's -(from -1) and also - to the right of 1. The right of 2 is positive. Does that mean that the relative min are -1 and 1, with the relative max being 2? Please help as soon as you can.

As you can see, g'(1)=0 twice. So, g' is negative on both sides. That means the function is falling, comes to a momentary stop, and then keeps on falling. (Think of y=x^3, y'=3x^2. y'=0, but is rising on both sides.)

g(1) is neither a max nor a min.

Since g' changes from + to - at x = -1, g rises, reaches a max, then falls.

Near x=2, g falls, reaches a min, and starts rising.

Thanks. That helped clear it up, because I wasn't sure what the squared would do. Also, a similar function of h''(x)=(x-1)(x+1)^2(x+2) asks what values of x h''(x) concaves up or down for, and like the other problem, I made a sign chart and got the same critical numbers with the same signs. Based off of the signs, does h''(x) concave down on (-1,1) and 1,2) and h''(x) concaves up on (-infinity, -1) and (2, infinity)?

correct.

Having trouble visualizing that, where h tries to change from concave down to up at x = -1, but fails, and stays concave down?

h(x) = x^6/30 + 3x^5/20 + x^4/12 - x^3/2 - x^2

It's a strange bird. You can see it straightening out, then staying concave down until x=1:

http://www.wolframalpha.com/input/?i=x%5E6%2F30+%2B+3x%5E5%2F20+%2B+x%5E4%2F12+-+x%5E3%2F2+-+x%5E2

To find the relative minimum and maximum of a function, you need to examine the behaviors of the function's derivative, g'(x), as you have done. A sign chart is a helpful tool for visualizing these behaviors. Let's analyze it step by step:

1. Start by finding the critical points of g'(x). These are the values of x where g'(x) is either zero or undefined. In this case, the derivative g'(x) = (x+1)(x-1)^2(x-2) has critical points at x = -1, x = 1, and x = 2.

2. Now, construct the sign chart by marking intervals on the x-axis corresponding to the spaces between these critical points.

3. Next, choose test points from each interval and plug them into g'(x) to determine the sign of the derivative.

4. With the values you tested, mark the signs in each interval on the chart.

Based on your description, it seems like you have already constructed the sign chart correctly. Let's recap what you mentioned:

- To the left of -1, g'(x) is positive. This means that the derivative is greater than zero in this interval.
- To the right of -1 and to the left of 1, g'(x) is negative. This indicates that the derivative is less than zero in this interval.
- To the right of 1 and to the left of 2, g'(x) is negative. Therefore, the derivative is also less than zero here.
- To the right of 2, g'(x) is positive. This tells us that the derivative is greater than zero in this interval.

Now, let's interpret these results:

- Where the derivative changes from positive to negative (to the left of -1 and between -1 and 1), g'(x) goes from increasing to decreasing. This suggests the presence of a relative maximum at x = -1.
- Similarly, where the derivative changes from negative to positive (between 1 and 2), g'(x) goes from decreasing to increasing. This implies the existence of a relative minimum at x = 1.
- Finally, as g'(x) is positive to the right of 2, it indicates that the function is increasing beyond x = 2. Therefore, x = 2 is not a relative maximum but rather a local minimum or turning point.

To summarize:

Relative maximum: x = -1
Relative minimum: x = 1
Local minimum/turning point: x = 2

It is worth noting that these are only the x-values of the relative minimum and maximum points. To find the corresponding y-values, you will need to substitute these x-values back into the original function g(x).