finding critical point involving arcsin

[calcnoob]

hi i have a certain calc problem im not sure if im doing this correctly im trying to find critical points

the equation is fx=2x^2+2cosx^-1

when i find the derivative i get 4x-(2/square root of (1-x^2))

i multiply the 4x by the square root of (1-x^2) to get the same denominator for both terms and i end up getting (-4x^2+4x-2)/square root of 1-x^2

i set the numerator to equal zero and this is where i get stuck at unless i did the entire thing wrong

 

[calcnoob]

woops i meant arccos not arcsin

 

[HallsofIvy]

hi i have a certain calc problem im not sure if im doing this correctly im trying to find critical points

the equation is fx=2x^2+2cosx^-1

Actually, what you have here is cos(1/x)!! The "-1" is misplaced:
you mean $\displaystyle f(x)= 2x^2+ 2cos^{-1}(x)$

when i find the derivative i get 4x-(2/square root of (1-x^2))

i multiply the 4x by the square root of (1-x^2) to get the same denominator for both terms and i end up getting (-4x^2+4x-2)/square root of 1-x^2

What became of the square root? What you should have is
$\displaystyle \frac{4x\sqrt{1- x^2}- 2}{\sqrt{1- x^2}}$

i set the numerator to equal zero and this is where i get stuck at unless i did the entire thing wrong

 

[calcnoob]

ok thanks but how do i solve the numerator when it is set to zero

4x(square root of 1-x^2)-2=0 im not sure how to go on from here

 

[HallsofIvy]

How about you just proceed the way you would with any equation- add 2 to both sides to get
$\displaystyle 4x\sqrt{1- x^2}= 2$

Now, get rid of the square root by, of course, squaring both sides.

 

[calcnoob]

ok i get 16x^2(1-x^2)=4

then i get -16x^3+16x^2-4=0

i factor it and get -4(4x^3-4x^2+1)=0