Question

find the values of x where the extreme values of the function y=x^3-7x-5x occur.

Answer 1

i hope this is calculus

Answer 2

and i hope there is a typo there too
check the question, because it looks funny

Answer 3

could it be \( y=x^3-7x^2-5x\) ?

Answer 4

no there is no typo its like this in the book and the answers choice i get is not there

Answer 5

probably a typo in the book then
why write \(x^3-7x-5x\) instead of \(x^3-12x\)

Answer 6

\[y=x^3-7x-5x\]

Answer 7

then start with
\[y=x^3-12x\] making \[y'=3x^2-12\] and set that equal to zero and solve

Answer 8

i get 2 and -2
but that is not my choices

Answer 9

(a) x=-0.33, x=5
(b) x=0.33, x=-5
(c) x=0.33, x=5
(d) x=2, x=5
(e) x=0.65 ,x=7.65

Answer 10

yeah so do i
that is why i think it is a typo

Answer 11

lets try
\[y=x^3-7x^2-5x\\
y'=3x^2-14x-5\\
y'=(3x+1)(x-5)\]

Answer 12

then you get critical points at \(x=-\frac{1}{3},x=5\)