Calculus- Early Transcendentals, 2021a

178 Applications of Derivatives
Perhaps remarkably, this special case is all we need to prove the more general one as well.
Theorem 5.22: Mean Value Theorem
Suppose that f (x) has a derivative on the interval (a,b) and is continuous on the interval [a,b]. Then
at some value c ∈ (a,b), f ′ f (b) − f (a)
(c)= .
b − a
f (b) − f (a)
Proof. Let m = , and consider a new function g(x)= f (x) − m(x − a) − f (a). We know that
b − a
g(x) has a derivative everywhere, since g ′ (x) = f ′ (x) − m. We can compute g(a) = f (a) − m(a − a) −
f (a)=0and
g(b)= f (b) − m(b − a) − f (a) =
f (b) − f (a)
f (b) − (b − a) − f (a)
b − a
= f (b) − ( f (b) − f (a)) − f (a)=0.
So the height of g(x) is the same at both endpoints. This means, by Rolle’s Theorem, that at some c,
g ′ (c)=0. But we know that g ′ (c)= f ′ (c) − m,so
which turns into
exactly what we want.
0 = f ′ (c) − m = f ′ (c) −
f ′ (c)=
f (b) − f (a)
,
b − a
f (b) − f (a)
,
b − a
♣
The Mean Value Theorem is illustrated below showing the existence of a point x = c for a function
f (x) where the tangent line at x = c (with slope f ′ (c)) is parallel to the secant line connecting A(a, f (a))
and B(b, f (b)) (with slope
f (b)− f (a)
b−a
):