Question

If f(x) is differentiable for the closed interval [-3, 2] such that f(-3) = 4 and f(2) = 4, then there exists a value c, -3 < c < 2 such that
f(c) = 0

f '(c) = 0

f (c) = 5

f '(c) = 5

Answer 1

@3mar

Answer 2

@phi

Answer 3

f() starts and ends at the same y-value
that means its average slope over the interval is zero
by some theorem we know that there must be at least one point where the instantaneous slope is also zero.

Answer 4

so f(c)=0? or would it be the derivative, f'(c)=0

Answer 5

@phi

Answer 6

they don't tell us any details about f(x), so they mean all f(x)'s have the property they are asking about.

one simple f(x) would be a straight line y= 4

Answer 7

Lagrange mean value theorem

Answer 8

there must be at least one point where the instantaneous slope is also zero.

how do you find the "slope" of a function ?

Answer 9

The slope is the derivative right? I know its the change in y over the change in x

Answer 10

yes, so we expect some value (call it c) where f' ( c) = 0

This question is asking about
https://en.wikibooks.org/wiki/Calculus/Rolle%27s_Theorem

Answer 11

and if (for example)
f(x) = 4
was the function, we see there is no c where f( c) = 0
so we can rule that out (even if we did not know about Rolle's Thm)

Answer 12

U may use the rolles theorem too.

If a function is continuous and differentiable in the closed interval [a,b] then there must be a point in the interval where f'(x)=\(\frac{f(b)-f(a)}{b-a}\)

Answer 13

so f(2)-f(-3)/-3-2 = 4-4/-5 = 0/-5= 0 so f'(x)=0

Answer 14

yes. Though that is called the "Mean Value Theorem" which includes Rolle's Thm
(Rolle's assumes f(a) = f(b) , i.e. assumes zero slope)

Answer 15

https://en.wikibooks.org/wiki/Calculus/Mean_Value_Theorem

I'm sure you will be asked about this one also.

Answer 16

Okay perfect thank you so much