Use mean value theorem

Question

𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥:

𝑓(𝑥) = {

𝑥

3 + 2𝑥

2 − 4𝑥 + 2, 𝑖𝑓 𝑥 ≤ 1

3𝑥 − 2, 𝑖𝑓 𝑥 > 1

𝐹𝑜𝑟 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 − 4 ≤ 𝑥 ≤ 2, 𝑓𝑖𝑛𝑑 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑐 𝑡ℎ𝑎𝑡 𝑠𝑎𝑡𝑖𝑠𝑓𝑦 𝑡ℎ𝑒 𝑀𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚

𝑀𝑒𝑎𝑛 𝑉𝑎𝑙𝑢𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 (𝐻𝑖𝑛𝑡: 𝑓(−4) = −14) [8]

2.2 𝑆𝑢𝑝𝑝𝑜𝑠𝑒 𝑓 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑎. 𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑎. [8]

Answer 1

This is a statement of the Mean Value Theorem:

"Let f: [a, b] → R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), where a < b. Then there exists some c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a)."

The given function f satisfies the conditions of the Mean Value Theorem.

It is continuous on the closed interval [-4, 2]

and differentiable on the open interval (-4, 2).

We know those statements to be true because the problem already tells us "the following function f which is differentiable for all x".

If we did not have that statement, we would have to reason that f is a piecewise function where each piece is a polynomial, thus each piece is continuous and differentiable. We would then have to do some additional work at x = 1.

For the given function f, we have:

f(-4) = (-4)³ + 2 * (-4)² - 4 * (-4) + 2

= -64 + 2 * 16 + 16 + 2

= -64 + 32 + 16 + 2

= -14

f(2) = 3 * 2 - 2 = 4

(f(2) - f(-4)) / (2 - (-4)) = (4 - (-14)) / (2 + 4)

= (4 + 14) / (2 + 4)

= 18 / 6

= 3

The derivative of the function f is given by:

f'(x) =

3x² + 4x - 4, if x ≤ 1

3, if x > 1

Thus, we immediately see that all the values of c between 1 and 2 satisfy the MVT.

Let us now investigate whether there are any values of c between -4 and 1 that also satisfy the MVT.

3x² + 4x - 4 = 3

⇔ 3x² + 4x - 4 - 3 = 3 - 3

⇔ 3x² + 4x - 7 = 0

⇔ (3x + 7)(x - 1) = 0

⇔ 3x + 7 = 0 or x - 1 = 0

⇔ x = -7/3 or x = 1

-9/3 < -7/3 < -6/3 so -3 < -7/3 < -2 so -7/3 is indeed in the interval (-4, 1]

Thus, the set of values of c that satisfy the MVT for the function f on the interval (-4, 2) is:

{-7/3} u [1, 2)

That is, the value -7/3 and all the real numbers between 1 (inclusive) and 2 (exclusive).

"Suppose f is differentiable at a. Show that f is continuous at a."

This is an extremely important result in calculus and its proof can easily be found in countless many sources.