Find c for the mean value theorem, f(x) = x^3 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 on [0, 3].

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 [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 = 0

And f of 3 is 27.

f(3) = 3³ = 27

2 The average slope

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

It is 9.

f(3) − f(0)3 − 0 = 27 − 03 − 0 = 273 = 9

Careful!

Divide by the width, 3 minus 0.

The average slope is 9, not 27.

Careful: not f(3) − f(0) = 27 ✗

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) = 9.

3c² = 9

Divide by 3.

c² = 3

Take the square root.

c is plus or minus √3, about 1.73.

c = √3 or −√3

4 Keep the ones inside

c, about 1.73, is between 0 and 3.

Keep it.

c = √3 ≈ 1.73 between 0 and 3 ✓

c, about −1.73, is not between 0 and 3.

It does not count.

c = −√3 ≈ −1.73 not between 0 and 3 ✗

So c = √3, about 1.73.

= c = √3 ≈ 1.73

5 Check

Let's check: put c into f′.

We get 9, the average slope.

Right!

f′(1.7321) ≈ 9

Let's look back.

First: check the conditions.

Then: the average slope.

Then: find c.

Then: keep the ones inside.

Last: check.

Answer