ummm cube root (1 Viewer)

[blackbunny]

i know how this function looks like but i have no idea how to get the poins out and skech it cuberoot(1-x^2)

 

[withoutaface]

Just need to find the points of contact with the axes (+/-1, 0)
y'= -2x/3(1-x^2)^2/3, so the turning point is at ( 0, 1), and going by that i think it would look like a circle, albeit a little mongelated and only the part above the x axis would be drawn.

 

[blackbunny]

yyes how did u get the turning point, u cant just say the derivative and give the turning point. so do i do this 1-x^2=0 >>>>>> x=0 >>>>y=0

 

[Will Hunting]

i think it would look like a circle, albeit a little mongelated and only the part above the x axis would be drawn

Nah, more like that scaaary Darth Vader helmet from Star Wars...
Remember that cuberoots of -ve numbers exist (eg. cubert(-8) = -2), so the graph can be drawn for f(x) < 0 (it is a function because the cubert of a number only ever has 1 real sol'n).

First, f(-x) = cubert[1 - (-x)^2] = cubert(1 - x^2) = f(x)
So, f(x) is even

If it's even, then we need only work with the graph on one side of the y-axis, after which we can reflect it in the y-axis. The positive side's the best:

x = 1, y = 0 an x-intercept (therefore, so is x = -1)
x = 0, y = 1 the y-intercept (and st. pt., from withoutaface's calculations)

Eq'n can be written, f(x) = cubert[(1-x)(1+x)]

Using this, and considering x < 1, f(x) becomes

cubert[(+ve)(+ve)] > 0 (This reads cube root of a positive times a positive)

Therefore, by symmetry about the y-axis, y > 0 for |x| < 1, or, in other words, f(x) is positive for the domain -1 < x < 1

This part of the graph's steeper than a regular semi-circle, though, because the cubert of a number < 0 is much bigger than the sqrt of the same number

Again, using f(x) = cubert[(1-x)(1+x)], consider x > 1,

cubert[(-ve)(+ve)] < 0

Therefore, y < 0 for |x| > 1, or f(x) is negative for the domain x > 1, x < -1

The curve is always decreasing for x > 0 and always increasing for x < 0

Couple of special pts: (3, -2); (-3, -2)

That should be enough to get you started!

 

[Trev]

This is a picture I just drew of it. Quite bad actually!