YAMT: Derivatives continued... Calculus work... Need help with 2nd derivative too

[Ricemarine]

f(x) = x^(2/3) * (x-4)^2

I'm trying to get f ' (x)...
So, using the product rule, I get...

f'(x) = 2/3x^(-1/3) * (x-4)^2 + x^(2/3) * 2 (x-4)

So to get it more simplified, I take out a 2/3x^(-1/3)

f ' (x) = 2/3x^(-1/3) [(x-4)^2 + 3x * (x-4)

Problem #1: Now apparently, I was told that you were not allowed to extract the 2/3 from the equation (by a tutor, not the teacher), which I don't know why, because then you screw yourself more making the problem more complicated than it should be...

f ' (x) = 2/3x^(-1/3) [(x^2-8x+16) + 3x^2 - 12x]

f ' (x) = 2/3x^(-1/3) [(4x^2 - 20x + 16]

f ' (x) = 2/3x^(-1/3) [(4x-4) (x-4)]

f ' (x) = 8/3x^(-1/3) [(x-1)(x-4)]

The critical points become 1 and 4...


Now for f ''(x)

f '' (x) = -8/9x^(-4/3) * (x^2-5x-4) + 8/3x^(-1/3) * (2x-5)

Problem #2: Was I supposed to add 2x outside of (2x-5)?...

Moving onward

f ''(x) = -8/9^(2/3) * 40/9x^(-1/3) * 32/9x^(-4/3) + 16/3x^(2/3) - 40/3x^(-1/3).

f '' (x) = -8/9x^(2/3) * 40/9x^(-1/3) [ 32/9x^(-4/3) + 6/5x - 15x ]

8/9x^(2/3) * 40/9x^(-1/3) [ 32/9x^(-4/3) - 69/5x ]

Problem #3: Did I do those calculations right? If not, where did I mess up? I'm supposed to find the inflection points from where the concavity changes...


Yeah, thank you so much for your help if you read this... and help me answer. I have tried math tutors, but they all went their own separate paths on how to approach this, and have no idea where to go.

 

[Ricemarine]

bump

 

[Legendary]

Edit: My explanation was wrong. Your problem is still with algebra though. One sec.

 

[Kyteland]

One easy way to check if you've done it right is to use this site: http://integrals.wolfram.com/index.jsp That way you can see if you've made any simple arithmetic errors.

You can extract anything you want from the equation. It'll still be the same equation. Doing that may or may not make reducing the problem easier. It's generally nice to keep whole numbers wherever you can, and taking out the 2/3 factor seemed to do that. Your answer for f'(x) seems to be correct.

for f"(x) the only issue I see is that it should be (x^2-5x+4) instead of (x^2-5x-4). After that it simplifies to f"(x) = 8/9*x^(-4/3) * (5*x^2-10*x-4) I don't know why you tried to simplify everything like you did, but usually you would try to pull out the stray power of 1/3 (or in this case 4/3) from your equation.

 

[Kyteland]

Originally posted by: Legendary
Your problem is not with calculus but with algebra - you can't remove any quantity of x^(-1/3) because it DOES NOT EXIST IN THE SECOND TERM. Say you have 3x + 2y, can you extract anything from there? Of course not, since x and y are not the same. Now change x to x^(-1/3) and y to x^(2/3) and you'll see your problem.

You need to go back and pay attention in math class. He did that exactly right.

x^(2/3) = x^(3/3)*x^(-1/3) = x*x^(-1/3)

 

[Ricemarine]

Originally posted by: Kyteland
One easy way to check if you've done it right is to use this site: http://integrals.wolfram.com/index.jsp That way you can see if you've made any simple arithmetic errors.

You can extract anything you want from the equation. It'll still be the same equation. Doing that may or may not make reducing the problem easier. It's generally nice to keep whole numbers wherever you can, and taking out the 2/3 factor seemed to do that. Your answer for f'(x) seems to be correct.

for f"(x) the only issue I see is that it should be (x^2-5x+4) instead of (x^2-5x-4). After that it simplifies to f"(x) = 8/9*x^(-4/3) * (5*x^2-10*x-4) I don't know why you tried to simplify everything like you did, but usually you would try to pull out the stray power of 1/3 (or in this case 4/3) from your equation.

How would you go about pulling that out?...
Damn product rules...

 

[Journer]

i'm doing the same stuff in my cal class and i friggin hate it...its the last math for me though...EVER HAHAHAHAHHAHA!!!!!!!!!!!

 

[Kyteland]

Originally posted by: Ricemarine

Originally posted by: Kyteland
One easy way to check if you've done it right is to use this site: http://integrals.wolfram.com/index.jsp That way you can see if you've made any simple arithmetic errors.

You can extract anything you want from the equation. It'll still be the same equation. Doing that may or may not make reducing the problem easier. It's generally nice to keep whole numbers wherever you can, and taking out the 2/3 factor seemed to do that. Your answer for f'(x) seems to be correct.

for f"(x) the only issue I see is that it should be (x^2-5x+4) instead of (x^2-5x-4). After that it simplifies to f"(x) = 8/9*x^(-4/3) * (5*x^2-10*x-4) I don't know why you tried to simplify everything like you did, but usually you would try to pull out the stray power of 1/3 (or in this case 4/3) from your equation.

How would you go about pulling that out?...
Damn product rules...

Fixing your small mistake, you already had this, which is right:
f '' (x) = -8/9x^(-4/3) * (x^2-5x+4) + 8/3x^(-1/3) * (2x-5)

"Pulling out" a factor is simple algebra and has nothing to do with the product rule. If you want to pull a factor of y out of your equation then multiply by y/y, which is equal to 1, and then simplify. So in your case you want to remove a factor of x^(-4/3)

f"(x) = -8/9x^(-4/3) * (x^2-5x+4) + 8/3x^(-1/3) * (2x-5)
= x^(-4/3)/x^(-4/3)*-8/9x^(-4/3) * (x^2-5x+4) + x^(-4/3)/x^(-4/3)*8/3x^(-1/3) * (2x-5) // multiply everything by x^(-4/3)/x^(-4/3), which is 1.
= x^(-4/3)*(-8/9x^(-4/3)/x^(-4/3) * (x^2-5x+4) + 8/3x^(-1/3)/x^(-4/3) * (2x-5)) // pull out your factor of x^(-4/3).
= x^(-4/3)*(-8/9 * (x^2-5x+4) + 8/3*x * (2x-5)) // simplify
= (8/9) * x^(-4/3) * (-(x^2-5x+4) + 3x*(2x-5)) // simplify some more
= (8/9) * x^(-4/3) * (-x^2+5x-4+6x^2-15x) // etc.
= (8/9) * x^(-4/3) * (5x^2-10x-4) // etc.

I did a lot of unnecessary steps in there just to be clear on what to do.

You applied the product rule just fine. What you got hung up on was the algebra required to reduce the equation to something simpler. That's the part you should be practicing.

 

[Ricemarine]

Yeah, I did not realize you could just factor out the -8/9x(-4/3)...

Thus it becomes -8/9x^(-4/3) [ -5x^2 + 10x + 4 ], which you can make switch the signs around so 5x^2 is positive...

So then quadratic formula gives 1+/- 3 root 2... But are those really the inflection points?... because f(1- 3 root 2) is 114... The other though is 4.66, which would make sense.

But thank you so much for your help thus far Kyteland.