f (x) is a differentiable function defined on the interval [0, 60].

x 0 15 30 45 60
f(x) -3 -1 2 6 2

Show that f '(c1) = 1/12 for some value x = c1 in the interval (0, 60). Justify your answer.

I'm thinking that the Mean Value Theorem applies here... but I have no clue as to how to explain or verify this.. please anyone help? I've been trying to figure this out for hours, really...

Question

f (x) is a differentiable function defined on the interval [0, 60].

x     0  15  30  45  60
f(x) -3  -1   2   6   2

Show that f '(c1) = 1/12 for some value x = c1 in the interval (0, 60). Justify your answer.

I'm thinking that the Mean Value Theorem applies here... but I have no clue as to how to explain or verify this.. please anyone help? I've been trying to figure this out for hours, really...

anonymous · Answer

Yes, you should use the mean value theorem

anonymous · Answer

There is c between a and b so that
f'(c)= (f(b)-f(a))/(b-a)

anonymous · Answer

yes

anonymous · Answer

i mean i know whawt mean value theorem is... do you think that's all i need to say on  this problem?

anonymous · Answer

(2-(-3)/(60-0) =5/60=1/12

anonymous · Answer

OH! i never thought of simplifying...

anonymous · Answer

a=0, b=60

anonymous · Answer

Wow. You are a genius! Thank you soooo much!!

anonymous · Answer

NP

anonymous · Answer

:D