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.
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.
3 answers
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
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
1 answer