Question

y = x −(256/x^2)
i need to fine the interecepts,realatiev extremas,point sof imflection and asymptote.

PleaSE any help would be apreciated

Answer 1

\[f(x)=x-\frac{256}{x^2}\]

Answer 2

\[f'(x)=1+\frac{512}{x^3}=\frac{x^3+512}{x^3}\]

Answer 3

first asymptotes. obviously one at x = 0 because this is undefined there. a slant asymptote at
\[y=x\]

Answer 4

ok now set the derivative = 0 and solve to get critical points. i.e. set
\[x^3+512=0\]
\[x^3=-512\]
\[x=-8\] that is your critical point

Answer 5

okk

Answer 6

it is a relative max

Answer 7

we know this because the derivative is positive to the left of -8 and negative to the right, which means your function was increasing and then decreasing. hence a max

Answer 8

to find inflection point take derivative again. don't use the quotient rule, start with
\[f'(x)=1+\frac{215}{x^3}\] derivative is
\[f''(x)=-\frac{1536}{x^4}\]

Answer 9

this thing is obviously negative for all values of x because the denominator is positive always and you have a minus sign out front, so function is always concave down

Answer 10

this also shows that -8 is a relative max, because the second derivative is negative there

Answer 11

and that is all except remember that -8 is not the relative max, it is the x value that gives the relative max. the relative max is what you get when you evaluate the function there. i will let you do that for yourself

Answer 12

if you want a nice picture of your function look here
http://www.wolframalpha.com/input/?i=y%3Dx-256%2Fx^2

Answer 13

thanks satelite, so what about the interecpt..r they none?and also theres no relative minimum then n only a maximum right?and to fine the maximum i would just plug -8 ninto the second derivitve right

Answer 14

thanks satelite, so what about the interecpt..r they none?and also theres no relative minimum then n only a maximum right?and to fine the maximum i would just plug -8 ninto the second derivitve right

Answer 15

no no, for maximum you want maximum of the function!

Answer 16

plug -8 into the function, not the derivative or the second derivative

Answer 17

yes there is no relative min. only one critical point

Answer 18

there is no y intercept since this has an asymptote at x = 0 (the y axis) it never touches it

Answer 19

the x intercept is where the function is 0. set it = 0 and solve

Answer 20

it is zero if\[ x = \sqrt[3]{256}\]