Mean value theorem for f(x) = x^3 - 3x^2 on [0, 3] Step by Step

Step by step

The question

Question: find the number c from the mean value theorem for f(x) = x³ − 3x² on [0, 3].

f(x) = x³ − 3x²

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, 3], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of 0 is 0.

f(0) = 0³ − 3(0)² = 0 + 0 = 0

And f of 3 is 0.

f(3) = 3³ − 3(3)² = 27 − 27 = 0

2 The average slope

The average slope from 0 to 3: the change in f over the change in x.

It is 0.

f(3) − f(0)3 − 0 = 0 − 03 − 0 = 03 = 0

Careful!

Divide by the width, 3 minus 0.

The average slope is 0, not 0.

Careful: not f(3) − f(0) = 0 ✗

3 Find c

The slope: use the power rule.

f′(x) = 3x^(3 − 1) − 3 × 2x^(2 − 1)

Now we have f′(x) = 3x² − 6x.

= 3x² − 6x

At c the slope is the average slope: f′(c) = 0.

3c² − 6c = 0

Take out the 3.

3(c² − 2c) = 0

We need two numbers that multiply to 0 and add to −2: 0 and −2.

3c(c − 2) = 0

0 × (−2) = 0, 0 + (−2) = −2

So c = 0 or c = 2.

c = 0 or c = 2

4 Keep the ones inside

c = 0 is not between 0 and 3.

It does not count.

c = 0 not between 0 and 3 ✗

c = 2 is between 0 and 3.

Keep it.

c = 2 between 0 and 3 ✓

So c = 2.

= c = 2

5 Check

Let's check: put c into f′.

We get 0, the average slope.

Right!

f′(2) = 0

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