Question

Find all the x-coordinates of all the critic. points & inflecton points in the interval (0,2pi) if y=(sin(x))^2 cosx.

I found first derivative = 2cosx^2sinx - sinx^3

Second derivative 2cosx^3-7cosxsinx^2

I know you can find your critical points by finding the zeros of the first derivative and the inflection points of the second derivative but I am completely stuck I would love for someone to walk me through it. Thanks!!

Answer 1

When I graphed in on a calculater there were a lot more answers than I was coming up with. I have to show all my work and give exact answers and I am just at a loss. :(

Answer 2

\[
y = \{\sin(x)\}^2\cos(x) ~~~\text{or}~~~y = \sin(x^2)\cos(x)
\]

Answer 3

the first one

Answer 4

\[
y = \sin^2(x)\cos(x) \\
y' = 2\sin(x)\cos(x)\cos(x) + \sin^2(x)(-\sin(x)) \\
y' = 2\sin(x)\cos^2(x) - \sin^3(x) = 0 \\
\sin(x) * (2\cos^2(x) - \sin^2(x)) = 0 \\
\sin(x) * (2 - 2\sin^2(x) - \sin^2(x)) = 0 \\
\sin(x) * (2 - 3\sin^2(x)) = 0 \\
\sin(x) = 0 ~~~\text{or}~~~\sin^2(x) = \frac 23
\]

Answer 5

\[
\sin(x) = 0 \implies x = 0, \pi, 2\pi \\
\text{But they want x in the OPEN interval }(0, 2\pi)~
\text{and so end points are excluded.} \\
x = \pi \\
\sin^2(x) = \frac 23 \implies \sin(x) = \sqrt{\frac 23} = 0.8165 \\
x = 0.9553, ~~ x = \pi - 0.9553 = 2.1863 \\
x = 0, ~~0.9553, ~~2.1863
\]

Answer 6

Actually when I took the square root I should consider both positive and negative values.
So we need to find x for sin(x) = -0.8165 too.

Answer 7

I was thinking the open interval would mean they were excluded, thanks for clarifying! He said he wanted complete exact answers and gave us some examples, would an exact answer be the same as sin (inverse) (2/3)?

Answer 8

I dont think I need to go into that much detail, he hates calculators and is allowing us to use calculators on this but he wants a very simple answer that is exact

Answer 9

Exact answer is:\[
x = \pi, ~~x=\sin^{-1}\left(\pm\sqrt{\frac23}\right)
\]

Answer 10

Oh yeah +/-

Answer 11

So there will be 5 critical points. x = pi is one.
sine inverse of +sqrt(2/3) will give one angle in the first quadrant and one angle in the second quadrant where the sine function is positive.
sine inverse of -sqrt(2/3) will give one angle in the third quadrant and one angle in the fourth quadrant where the sine function is negative.

Answer 12

THANK YOU!! Also what about the inflection points?

Answer 13

I am just having a tough time going through these :/

Answer 14

\[
y' = 2\sin(x)\cos^2(x) - \sin^3(x) \\
y'' = 2\sin(x) * 2\cos(x)*(-\sin(x)) + 2\cos^2(x)(\cos(x)) - 3\sin^2(x)(\cos(x)) = 0 \\
-4\sin^2(x)\cos(x) + 2\cos^3(x) - 3\sin^2(x)\cos(x) = 0 \\
-7\sin^2(x)\cos(x) + 2\cos^3(x) = 0 \\
\cos(x)(2\cos^2(x) - 7\sin^2(x)) = 0 \\
\cos(x)(2 - 9\sin^2(x)) = 0 \\
\cos(x) = 0 ~~\text{or}~~\sin^2(x) = \frac 29 \\
\cos(x) = 0 ~~\text{or}~~\sin(x) = \pm\frac {\sqrt{2}}{3} \\
x = \frac{\pi}{2}, ~~x = \frac{3\pi}{2}, ~~x = \sin^{-1}\left( \pm\frac {\sqrt{2}}{3} \right)
\]

Answer 15

Thanks so much!