Question

If f(x) is differentiable for the closed interval [−1, 4] such that f(−1) = −3 and f(4) = 12, then there exists a value c, −1< c < 4 such that

a)f'(c)=3
b)f'(c)=0
C)f(c)=-15
d)f(c)=3

I understand that you are supposed to use the mean value theorem, but i dont understand how. If you could walk me through it step-by-step, that would be greatly appreciated.

Answer 1

if you google MVT you will see diagrams of what it means. All it amounts to is that if the conditions are met, there will be one or more places in the interval where the tangent line is parallel to the line joining the points at the ends of the interval.

In this case, the line joining the points on the graph at the ends of the interval has slope

(f(4)-f(-1))/(4-(-1))
= (12-(-3))/(4-(-1))
= 15/5
= 3

So, somewhere in [-1,4], the tangent line will have a slope of 3.

That is choice (a)

Answer 2

the function climbs uphill 15 from -3 to +12
in its jaunt of +5 from -1 to +4
that means that somewhere in between there the slope of the function is

slope = 15/5 = 3
so at some x = c between x = -1 and x=+4slope = f'(c) = 3
so
a)