Find c for the mean value theorem, f(x) = x^3

Step by step

The question

Question: find the number c from the mean value theorem for f(x) = x³ − x on [0, 2].

f(x) = x³ − x

The theorem: for some c between a and b, the slope at c equals the average slope from a to b.

f′(c) = f(b) − f(a)b − a

for some c between a and b

1 Check the conditions

f is a polynomial.

Its graph has no breaks and no sharp corners.

So it is continuous on [0, 2], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of 0 is 0.

f(0) = 0³ − 0 = 0 + 0 = 0

And f of 2 is 6.

f(2) = 2³ − 2 = 8 − 2 = 6

2 The average slope

The average slope from 0 to 2: the change in f over the change in x.

It is 3.

f(2) − f(0)2 − 0 = 6 − 02 − 0 = 62 = 3

Careful!

Divide by the width, 2 minus 0.

The average slope is 3, not 6.

Careful: not f(2) − f(0) = 6 ✗

3 Find c

The slope: use the power rule.

f′(x) = 3x^(3 − 1) − 1

Now we have f′(x) = 3x² − 1.

= 3x² − 1

At c the slope is the average slope: f′(c) = 3.

3c² − 1 = 3

Add 1 on both sides.

3c² = 4

Divide by 3.

c² = 43

Take the square root.

c is plus or minus 2 over √3, about 1.15.

c = 2√3 or −2√3

4 Keep the ones inside

c, about 1.15, is between 0 and 2.

Keep it.

c = 2√3 ≈ 1.15 between 0 and 2 ✓

c, about −1.15, is not between 0 and 2.

It does not count.

c = −2√3 ≈ −1.15 not between 0 and 2 ✗

So c = 2 over √3, about 1.15.

= c = 2√3 ≈ 1.15

5 Check

Let's check: put c into f′.

We get 3, the average slope.

Right!

f′(1.1547) ≈ 3

And in a picture.

The line through the two ends has the average slope.

At c, the tangent line is parallel to it: the same slope.

Right!

Let's look back.

First: check the conditions.

Then: the average slope.

Then: find c.

Then: keep the ones inside.

Last: check.

Answer