Mean Value Theorem: Verify that f(x)=sin(x) on [0,π/2) satisfies the hypothesis

[akarbarz]

Verify that f(x)=sin(x) on [0,π/2) satisfies the hypothesis of the Mean Value Theorem on the interval and find all C values that satisfy the conclusion.
So far I have:

- sin(x) is continuous and differentiable on the interval, therefore the Mean Value Theorem is satisfied

So,
f(b)-f(a)/b-a = f'c
(1-0)/(π/2-0)=cos(c)
(2/π) = cos(c) <------ this is where I am stuck, and not sure what to do next.

Thank you in advance for your help.

 

[ksdhart2]

What has your class covered so far about inverse trig functions? Specifically, what would happen if you applied the inverse cosine function to both sides of your equation?

\(\displaystyle cos^{-1} \left( \dfrac{2}{\pi} \right) = cos^{-1}(cos(c))\)

Where does that lead you? Be sure to remember that you've restricted the domain to (0, pi/2).

 

[tkhunny]

...and please work on your notation. I doubt you mean f(b) - f(a) / b-a. I'm guessing you mean [f(b)-f(a)]/[b-a]. Inline typing is NOT the same as textbook typesetting.

 

[akarbarz]

What has your class covered so far about inverse trig functions? Specifically, what would happen if you applied the inverse cosine function to both sides of your equation?

\(\displaystyle cos^{-1} \left( \dfrac{2}{\pi} \right) = cos^{-1}(cos(c))\)

Where does that lead you? Be sure to remember that you've restricted the domain to (0, pi/2).

Correct me if I am wrong but then would my answer just be c = cos^-1(2/π)
or I could say x is approximately .880689 ?
Thanks.

 

[ksdhart2]

Correct me if I am wrong but then would my answer just be c = cos^-1(2/π)
or I could say x is approximately .880689 ?
Thanks.

Yes, that's correct. But, in the future, if you're ever unsure about an answer, you can always check it yourself. In this case, you want to verify that ~0.880689 is the inverse cosine of 2/pi. So, take the cosine of that value, and you'll see that you, as expected, get 2/pi.

 

[akarbarz]

Yes, that's correct. But, in the future, if you're ever unsure about an answer, you can always check it yourself. In this case, you want to verify that ~0.880689 is the inverse cosine of 2/pi. So, take the cosine of that value, and you'll see that you, as expected, get 2/pi.

Ah yes, okay. Thank you very much! I appreciate the help.