anonymous
  • anonymous
Consider f(x)=sin^2(x)-cos(x) on the interval 0 is less than or equal to x which is less than or equal to pi. Determine the local extrema of f(x).
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[f(\theta)=\sin^{2}(\theta)-\cos(\theta)\] \[0\le \theta \le \pi\]
anonymous
  • anonymous
I know one of the local extrema is at x=0. There are two more.
anonymous
  • anonymous
in order to find the local extrema first differentiate f(x) with respect to x and then put f '(x) = 0 this will help u in getting local extrema

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I get:\[\sin(\theta)(2\cos(\theta)+1)=0\]
anonymous
  • anonymous
I know the sin (theta) piece is what give the local extrema at x=0. But I can't figure out the cos part.
anonymous
  • anonymous
ok 2{cos (theta) + 1} = 0 now can we write cos(theta) + 1 = 0?
anonymous
  • anonymous
cos(theta) = -1 cos(theta) = cos(n* pi) where n is an odd integer is this step clear to u ?
anonymous
  • anonymous
i m sorry its wrong
anonymous
  • anonymous
Yes, I am trying to find the angle that gives the odd integer?
anonymous
  • anonymous
let me correct its 2cos(theta) +1 = 0 2cos(theta) = -1 cos(theta) = -1/2 is this step clear to u?
anonymous
  • anonymous
as we know cosine function is negative in second and third quadrant therefore , value of theta must lie in pi/2 < theta < 3*pi/2
anonymous
  • anonymous
2pi/3 or 4 pi/3
anonymous
  • anonymous
yup
anonymous
  • anonymous
Do I then restrict those answers based on the domain?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
So then 2pi/3 is the only one that works?

Looking for something else?

Not the answer you are looking for? Search for more explanations.