Some calculus questions...?

Question

Hi...We are going over applic. of derivatives in class right now...I've got a trig. question and another one pertaining to Mean Value Theorem...To be quite honest, I'm not good with trig. functions and I don't understand the second question...

1. Find the function wiht the given derivative whose graph passes through P.
csc (theta) cot (theta) -3, P(pi/2, 0)
I got the answer csc (theta)-3(theta) +1 + 3pi/2
Please correct if wrong, and explain how you did so...

And...The second question...
2. Determine whether the function sastisfies the hypothesis of the Mean Value Theorem for the given interval.
f(x)=x^(1/3) [-1,5]

And like I said, I'm not really sure how to do this one...I would think that it doesn't satisfy the theorem, because the interval doesn't contain a 0...I'm probably wrong...PLEASE JUST HELP...

THANKS...

Answer 1

csc (theta) cot (theta) -3, P(pi/2, 0)

∫[cscθ cotθ - 3]dθ
= -cscθ - 3θ + c

When θ = π/2 expression = 0
ie -cscπ/2 - 3π/2 + c = 0
So c = 3π/2 + 1

Thus particular integral sought is -cscθ - 3θ + 3π/2 + 1

Mean Value Theorem states that in [a, b] there exists c such that
f'(c) = [f(b) - f(a)]/[b - a] if f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b)

f(x) = x^⅓; a = -1, b = 5
[f(5) - f(-1)]/[5 - -1]
= [5^⅓ + 1]/6

f'(x) = 1/[3x^⅔]

So f'(c) = 1/[3c^⅔]

Now solve 1/[3c^⅔] = [5^⅓ + 1]/6
So 3c^⅔ = 6/[5^⅓ + 1]
ie c^⅔ = 2/[5^⅓ + 1]
c ≈ 0.6340 which is in the interval [-1, 5]

However while f(x) is continuous on the closed interval [-1, 5] it is NOT differentiable at 0 which lies on the open interval (-1, 5) so the theorem falls over on this count

Answer 2

The first one is almost right. An antiderivative of
csc(theta) cot (theta) is -csc (theta).
So all you need is a minus sign in front of your
first term.
For the second one, what are the hypotheses(PLURAL!)
of the mean value theorem?
First, the function must be continuous on [a,b]. Satisfied.
Then the derivative of the function must exist on (a,b).
Does it?
Well, the derivative of x^(1/3) is 1/3*x^(-2/3),
which does not exist at 0. So the answer to
your second question is "NO".

Answer 3

This lim is inderterminate (zero/zero) however you will have precluded using L'Hopital's Rule. So..... e^x = a million + x + x^two/two + ........ As x is going to zero then e^x is going to a million + x Therefore your prohibit turns into lim (a million + h - a million)/3h = a million/three lim a million = a million/three with all of the above limits as h has a tendency to zero. Go in peace....