1. f(x) is increasing at the intervals (-2,0)U(2,0)
f(x) is decreasing at the intervals (-infinti, -2)U(2, infinti)
where f '(x)>0 f is increasing & f '(x)<0function is decreasing
2. at x= +/- 1.25 local max. the slope increases then decreases so they have to be max's.
3.concave up at (-infinti, 2) concave down (2, infinti)
we know where f is increasing up and then decreasing down determining the concavities.
4.quadratic maybe end behavior of x to the power of 5.
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i saw everything differently from you. but im not sure if im right. i agree wiht number 1 but not quite number 3. i think to see whether f(x) concaves up or down.. i think we have to look at whether the slope of f '(x) is positive or negative. so i think it concaves up where the slope of f '(x) is positive and concaves down where the slope of f '(x) is negative.
ReplyDeleteconcaves up: (-∞,-1.25)U(1.25,∞)
concaves down: (-1.25,0)U(1.25,2)
i dont know if im right or wring!1 i still kinda confused!!!
I agree w/ Wendy. Your number 1 looks great but number 3 needs help.
ReplyDeleteWith 2, that was my fault. I was asking for the extrema of f(x), not this graph, f'(x). If I was, you are still missing the local min at x=0.
4. No "end-behavior" here. This is strictly talking about this picture that you see. Quadratics are always to the degree of 2, not 4. 4th degrees are called quartics.
im losing it i think i lost count after some posts
ReplyDelete