Question

Find any relative extrema of the function. (Round your answers to three decimal places.)
f(x) = arcsec(x) − 3x
relative maximum (x, y) = ( )
relative minimum (x, y) = ( )

Answer 1

can someone please help me..

Answer 2

\[f(x)=\sec^{-1}(x)-3x\]
\[f'(x)=\frac{1}{x\sqrt{x^2-1}}-3\]

Answer 3

set this beast equal to zero and solve

Answer 4

cana u break it down for me till the simple algebra so i can solve it from there.please

Answer 5

ok if i can. is the derivative ok?

Answer 6

\[f'(x)=\frac{1}{x\sqrt{x^2-1}}-3=0\]
\[\frac{1}{x\sqrt{x^2-1}}=3\]
\[3x\sqrt{x^2-1}=1\]
\[9x^2(x^2-1)=1\]
\[9x^4-9x^2-1=0\]

Answer 7

now you have a quadratic equation is
\[x^2\] use the quadratic formula and solve for z in
\[9z^2-9z-1=0\]

Answer 8

or just cheat. may be easier
http://www.wolframalpha.com/input/?i=y%3Darcsec%283x%29-3x

Answer 9

satelitte i clciked on the link but what would be my relative maximum and relative minimum...could you point that out to me

Answer 10

it is at the very bottom of the link

Answer 11

max at x = stuff
min at x = stuff

Answer 12

but doesnt it have to be a point like (x,y) so that solves for x what about y.

Answer 13

sorry to be such a pain but could you help me get two a solid point for the min and maximum

Answer 14

oh wait i think i got it

Answer 15

ugh you plugged in 3x instead of just x into wolfram so i got it wrong its oki though

Answer 16

could u help me with something esle

Answer 17

but is said 3x in your problem yes?

Answer 18

i just used your original equation

Answer 19

http://www.wolframalpha.com/input/?i=f%28x%29%3Darcsec%28x%29-3x
i typed in just as you wrote it