Title | Mean Value Theorem |
---|---|
Course | Mathematics 1A |
Institution | University of New South Wales |
Pages | 1 |
File Size | 61.2 KB |
File Type | |
Total Downloads | 92 |
Total Views | 165 |
Mean Value Theorem...
Examples (note that, we are not usually interested in finding the 𝑐 explicitly, just knowing it exists is enough) •
Suppose that 𝑓: [1,4] → ℝ is given by 𝑓(𝑥) = 3√𝑥 − 4𝑥. Find a number 𝑐 𝜖 (1,4) that satisfies the conclusion of the MVT for 𝑓 on [1,4]. See immediately, the function is differentiable along the whole closed interval.
•
Using MVT to prove inequalities •
Use the MVT to show that √𝑥 + 4 < 2 + 𝑥 √𝑥 + 4 − 2 − < 0 4 Proof: fix 𝑥 > 0. Define function:
𝑥 4
for ∀𝑥 > 0. (Thurs March 14)
𝑡
𝑓(𝑡) = √𝑡 + 4 − 2 − on [0, 𝑥] 4
Note: 𝑓(0) = 0 *. Use MVT: check that 𝑓 is differentiable at any point of [0, 𝑥]. 𝑡 𝑡↦ 2+ differentiable on [0, 𝑥] 4
𝑡 ↦ √𝑡 + 4 differentiable on [0, 𝑥] 𝑡 differentiable on [0, 𝑥] 𝑡 ↦ √𝑡 + 4 − (2 + 4) These are elementary functions, therefore are differentiable. Therefore, we can use the MVT. By the MVT:
∴∗
𝑓(𝑥) 𝑥
= 𝑓′(𝑐) for a certain 0 < 𝑐 < 𝑥
𝑓(𝑥)−𝑓(0) 𝑥−0
= 𝑓′(𝑐) with 0 < 𝑐 < 𝑥.
Since 𝑥 > 0, we have that
𝑓(𝑥) 𝑥
< 0 ⟺ (if and only if) 𝑓(𝑥) < 0
In order to finish the proof, it is sufficient to show that 𝑓 ′ (𝑐) < 0. 𝑡 𝑓(𝑡) = √𝑡 + 4 − 2 − 4 1 1 𝑓 ′ (𝑡) = − 2√𝑡 + 4 4 =
1 𝑡
2√4( +1) 4
=4 ( 1
1 𝑡
√ +1 4
−
1
4
− 1)
since √ + 1 > 1, then 𝑓 ′ (𝑡) < 0 for any 𝑡 > 0 𝑡
4
In particular, 𝑓 ′ (𝑐) < 0. Since
𝑓(𝑥) 𝑥
= 𝑓′(𝑐) and 𝑥 > 0, we have 𝑓(𝑥) < 0....