Mean value theorem for f(x) = x^3 on [-1, 2] Step by Step

Step by step

The question

Question: find the number c from the mean value theorem for f(x) = x³ on [−1, 2].

f(x) = x³

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 [−1, 2], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of −1 is −1.

f(−1) = (−1)³ = −1

And f of 2 is 8.

f(2) = 2³ = 8

2 The average slope

The average slope from −1 to 2: the change in f over the change in x.

It is 3.

f(2) − f(−1)2 − −1 = 8 − (−1)2 − (−1) = 93 = 3

Careful!

Divide by the width, 2 minus −1.

The average slope is 3, not 9.

Careful: not f(2) − f(−1) = 9 ✗

3 Find c

The slope: use the power rule.

f′(x) = 3x^(3 − 1)

Now we have f′(x) = 3x².

= 3x²

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

3c² = 3

Take away 3 on both sides, so they stay equal.

3c² − 3 = 0

Take out the 3.

3(c² − 1) = 0

x² − 1 is a difference of two squares: x² minus 1 squared.

So it is (x − 1)(x + 1).

3(c + 1)(c − 1) = 0

a² − b² = (a − b)(a + b), b = 1

So c = −1 or c = 1.

c = −1 or c = 1

4 Keep the ones inside

c = −1 is not between −1 and 2.

It does not count.

c = −1 not between −1 and 2 ✗

c = 1 is between −1 and 2.

Keep it.

c = 1 between −1 and 2 ✓

So c = 1.

= c = 1

5 Check

Let's check: put c into f′.

We get 3, the average slope.

Right!

f′(1) = 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