Question

Calculus derivative problem.Help plz!!!!!

Answer 1

Show the steps for graphing y= (3 * x^(2/3))+x

Answer 2

I have tried it but failed multiple times

Answer 3

I have found the first derivative,then drew the sign chart,then second derivative,then sign chart again,then concavity but my graph wasnt simlar to graphing calculator results

Answer 4

Is that \(\large y=3x^{2/3}+x\)? If you're trying to graph something using derivatives, you'll probably want to start by finding when the first derivative is zero to find critical points, then plug in critical points to find critical values. Sounds like you've done that, can you show what you got so far so we can see what went wrong?

Answer 5

Yes.Plz shpw every step so I can see where I am making the error

Answer 6

Well, step one, take the derivative. What did you get for the first derivative?

Answer 7

I got y=1+2x^(-1/3)

Answer 8

then I found -8 and 0 to be critical points if I am right

Answer 9

Right. So then you set y=0 to find the critical points, so \(\Large\dfrac{2}{\sqrt[3]x}+1=0, \dfrac{2}{\sqrt[3]x}=-1, \sqrt[3]x=-2, x=-8\).
And yes, 0 is also a point worth looking at because the first derivative is undefined at x=0. So, we plug those critical points into the original equation to find critical values, right? And then we can probably take the second derivative to see if we can get some information about concavity?

Answer 10

I have done that . I got increasing for (infinity,8],decreasing fro [-8,0] and increasing [0,infinty). Local min (0,0) and local max (-8,4)

Answer 11

Then I found second derivative to be y**=(-2/3).(x^(-4/3))

Answer 12

then I found 0 to be the crtical point for 2nd fucntion if I am right

Answer 13

?

Answer 14

Sorry, working on it, trying to help on a few different problems at the same time.

Answer 15

haha-Ok-In case my computer runs out of battery,plz write the rest of solution for me with the graph.Thanks!I became a fan of you ;)

Answer 16

Are you stuck? Hopefully not ;)

Answer 17

Yeah, I'm basically getting the same answers as you. Also, the function is always concave, except for at 0, because 0 is a weird point. It's not a discontinuity because its limit is 0, but the first and second derivatives are both undefined at 0. Anyway, I'm just getting an always concave line, increasing until -8,4, then decreasing to 0,0, then increasing again to infinity. Weird function though and I'm rusty on this stuff so I'm not sure.

Answer 18

so how do you graph it

Answer 19

also concave up or down?

Answer 20

That's a crummy drawing, but something like that. Concave means concave down (iirc that they call n-shaped concave down), 'concave up' is called convex, math teachers are often bad at communicating that.