The Mean Value Theorem-1, guaranteed and verified 100% Pass

1


The Mean Value Theorem



Def. Let 𝑓 be a real valued function defined on a metric space 𝑋. We say that 𝑓
has a local maximum at a point 𝑝𝜖𝑋 if there exists a 𝛿 > 0 such that
𝑓(𝑞) ≤ 𝑓(𝑝) for all 𝑞𝜖𝑋 with 𝑑𝑋 (𝑝, 𝑞 ) < 𝛿 .
We say that 𝑓 has a local minimum at a point 𝑝𝜖𝑋 if there exists a 𝛿 > 0 such
that 𝑓(𝑝) ≤ 𝑓(𝑞) for all 𝑞𝜖𝑋 with 𝑑𝑋 (𝑝, 𝑞 ) < 𝛿 .




𝑦 = 𝑓(𝑥)

𝑝1 𝑝2

Local Max Local Min



Theorem: Let 𝑓: [𝑎, 𝑏] → ℝ. If 𝑓 has a local maximum or minimum at a point
𝑝𝜖(𝑎, 𝑏), and if 𝑓 ′ (𝑝) exists, then 𝑓 ′ (𝑝) = 0.

Proof: Suppose 𝑓 ′ (𝑝) exists and 𝑓(𝑝) is a local maximum.

𝑦 = 𝑓(𝑥)


𝑝 − 𝛿( 𝑝 )𝑝 + 𝛿



Then by the definition of a local maximum, there exists a 𝛿 > 0 such that

𝑓(𝑥) ≤ 𝑓(𝑝) for all 𝑥𝜖[𝑎, 𝑏] with |𝑥 − 𝑝| < 𝛿.

, 2


|𝑥 − 𝑝| < 𝛿
−𝛿 < 𝑥 − 𝑝 < 𝛿
𝑝 − 𝛿 < 𝑥 < 𝑝 + 𝛿.


Suppose that we take a point 𝑡, 𝑝 − 𝛿 < 𝑡 < 𝑝 , then we have:

𝑓(𝑡) − 𝑓(𝑝) ≤ 0 since 𝑓(𝑥) ≤ 𝑓(𝑝) for all 𝑥𝜖[𝑎, 𝑏] with |𝑥 − 𝑝| < 𝛿 and

𝑡−𝑝 <0 since 𝑡 < 𝑝;


So we have:

𝑓(𝑡 )−𝑓(𝑝) 𝑓(𝑡 )−𝑓(𝑝)
≥ 0; for 𝑡 < 𝑝. Thus we can say: lim− ≥ 0.
𝑡−𝑝 𝑡→𝑝 𝑡−𝑝



Now suppose we take a point , 𝑝 < 𝑡 < 𝑝 + 𝛿 .

So we have:

𝑓(𝑡) − 𝑓(𝑝) ≤ 0 since 𝑓(𝑥) ≤ 𝑓(𝑝) for all 𝑥𝜖𝑋 with |𝑥 − 𝑝| < 𝛿

𝑡−𝑝 >0 since 𝑝 < 𝑡;


So we have:

𝑓(𝑡 )−𝑓(𝑝) 𝑓(𝑡 )−𝑓(𝑝)
≤ 0; for 𝑡 > 𝑝. That gives us: lim+ ≤ 0.
𝑡−𝑝 𝑡→𝑝 𝑡−𝑝



Since 𝑓 ′ (𝑝) exists we must have:

𝑓(𝑡 )−𝑓(𝑝) 𝑓(𝑡 )−𝑓(𝑝)
0 ≤ lim− = lim+ ≤0
𝑡→𝑝 𝑡−𝑝 𝑡→𝑝 𝑡−𝑝

, 3


𝑓(𝑡 )−𝑓(𝑝)
Thus 𝑓 ′ (𝑝) = lim = 0.
𝑡→𝑝 𝑡−𝑝

A similar argument works when 𝑝 is a local minimum.



The next theorem will be used later to prove L’Hopital’s rule.



Theorem (Generalized Mean Value Theorem): If 𝑓, 𝑔: [𝑎, 𝑏] → ℝ, are
continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏), then there exists a point
𝑐𝜖(𝑎, 𝑏) at which:
[𝑓(𝑏) − 𝑓(𝑎)]𝑔′ (𝑐) = [𝑔(𝑏) − 𝑔(𝑎)]𝑓 ′ (𝑐).
Note: we could also write this result as:

𝑓(𝑏 )−𝑓(𝑎) 𝑓′ (𝑐)
= 𝑔′ (𝑐) ; This in turn could be written as:
𝑔(𝑏 )−𝑔(𝑎)

𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎 𝑓′ (𝑐)
𝑔(𝑏)−𝑔(𝑎) = i.e.
𝑔′ (𝑐)
𝑏−𝑎

𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑓 𝑜𝑣𝑒𝑟 [𝑎,𝑏] 𝐼𝑛𝑠𝑡. 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑓 𝑎𝑡 𝑐
= 𝐼𝑛𝑠𝑡. .
𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑔 𝑜𝑣𝑒𝑟 [𝑎,𝑏] 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑔 𝑎𝑡 𝑐



Proof: Let ℎ(𝑥) be defined by:

ℎ(𝑥 ) = [𝑓(𝑏) − 𝑓 (𝑎)]𝑔(𝑥 ) − [𝑔(𝑏) − 𝑔(𝑎)]𝑓(𝑥 ); 𝑎 ≤ 𝑥 ≤ 𝑏.
ℎ(𝑥) is continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏) because 𝑓(𝑥) and 𝑔(𝑥)
are.