Question

"Take the derivative, and solve for x. The equation:
y = (x^3) - (26^2) + (160x)"

Any and all help is greatly appreciated! :)

Answer 1

I got x = 4, & x = 13.33, but my book has something different...

Answer 2

Is it asking for dy/dx or d/dx ?

Answer 3

If we take the derivative with just respect to x, y is treated as a constant.

Answer 4

...it didn't specify, just the generic derivative of the equation, but no u substitution or anything.

Answer 5

Ok.

Answer 6

y = (x^3) - (26^2) + (160x)
y' = 3x^2 + 160

Answer 7

No -52x ?

Answer 8

i am assuming that is a typo right?

Answer 9

you put - (26^2)

Answer 10

\[y = x^3 - 26x^2+ 160x\]

Answer 11

if it is a typo then
y' = 3x^2 - 52x + 160

Answer 12

Okay, that's what I was getting too.

Answer 13

And x = 4 & 13.33, right??

Answer 14

(Used quadratic formula)

Answer 15

Set y' = 0
so
0 = 3x^2 - 52x + 160
(x-4)(3x-40)
x = 4, x = 40/3
yeah

Answer 16

Alright... Sweet. *God-dang book...

Answer 17

Thank you ! :))

Answer 18

Yeah, sometimes the book answers are wrong, and you're welcome. :)