Can someone please check my work (Relative Extremes)

[paulxzt]

Find all relative extrema of y = (2x) / (x+4)^3

I first got the y' = [ (x+4)^3 (2) - 2x(3(x+4)^2) ] / (x+4)^6

y' = [ 2(x+4)^3 -2x (3(x+4)^2) ] / (x+4)^6

y' = ( 2 - 6x ) / (x+4)

y' = 0 when x = 1/3. y' = dne at x = -4

then i found local min @ x = -4 and local max @ x = 1/3

can someone please tell me if this is right? Does relative extrema mean just the local min and max?
Also, is my algebra correct on the derivative? Please tell me if i made a mistake because i was never good at algebra. (Was the canceling x+4 correct?)

Thanks for any help

 

[stapel]

y' = ( 2 - 6x ) / (x+4)

I'm not seeing how you arrived at this expression for your first derivative...? You had:

. . . . .\(\displaystyle \L y'\, =\, \frac{2(x\, +\, 4)^3\, -\, 2x(3)(x\, +\, 4)^2}{((x\, +\, 4)^3)^2}\)

. . . . .\(\displaystyle \L y'\, =\, \frac{(x\, +\, 4)^2 \, (2(x\, +\, 4)\, -\, 6x)}{(x\, +\, 4)^6}\)

. . . . .\(\displaystyle \L y'\, =\, \frac{2x\, +\, 8\, -\, 6x}{(x\, +\, 4)^4}\)

I'm not seeing how you made the jump from the last line above to "(2 - 6x) / (x + 4)"...?

Thank you.

Eliz.

 

[paulxzt]

y' = ( 2 - 6x ) / (x+4)

I'm not seeing how you arrived at this expression for your first derivative...? You had:

. . . . .\(\displaystyle \L y'\, =\, \frac{2(x\, +\, 4)^3\, -\, 2x(3)(x\, +\, 4)^2}{((x\, +\, 4)^3)^2}\)

. . . . .\(\displaystyle \L y'\, =\, \frac{(x\, +\, 4)^2 \, (2(x\, +\, 4)\, -\, 6x)}{(x\, +\, 4)^6}\)

. . . . .\(\displaystyle \L y'\, =\, \frac{2x\, +\, 8\, -\, 6x}{(x\, +\, 4)^4}\)

I'm not seeing how you made the jump from the last line above to "(2 - 6x) / (x + 4)"...?

Thank you.

Eliz.

y' = [ 2(x+4)^3 -2x (3(x+4)^2) ] / (x+4)^6
I think I did it wrong but is it not correct to cancel out (x+4)^3 and (x+4)^2 from the top with the bottom?

 

[stapel]

...s it not correct to cancel out (x+4)^3 and (x+4)^2 from the top with the bottom?

You can only cancel factors, not terms.

Eliz.