Mean value theorem for f(x) = x^3 - 12x on [0, 4] Step by Step

Step by step

The question

Question: find the number c from the mean value theorem for f(x) = x³ − 12x on [0, 4].

f(x) = x³ − 12x

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, 4], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of 0 is 0.

f(0) = 0³ − 12(0) = 0 + 0 = 0

And f of 4 is 16.

f(4) = 4³ − 12(4) = 64 − 48 = 16

2 The average slope

The average slope from 0 to 4: the change in f over the change in x.

It is 4.

f(4) − f(0)4 − 0 = 16 − 04 − 0 = 164 = 4

Careful!

Divide by the width, 4 minus 0.

The average slope is 4, not 16.

Careful: not f(4) − f(0) = 16 ✗

3 Find c

The slope: use the power rule.

f′(x) = 3x^(3 − 1) − 12 × 1

Now we have f′(x) = 3x² − 12.

= 3x² − 12

At c the slope is the average slope: f′(c) = 4.

3c² − 12 = 4

Add 12 on both sides.

3c² = 16

Divide by 3.

c² = 163

Take the square root.

c is plus or minus 4 over √3, about 2.31.

c = 4√3 or −4√3

4 Keep the ones inside

c, about 2.31, is between 0 and 4.

Keep it.

c = 4√3 ≈ 2.31 between 0 and 4 ✓

c, about −2.31, is not between 0 and 4.

It does not count.

c = −4√3 ≈ −2.31 not between 0 and 4 ✗

So c = 4 over √3, about 2.31.

= c = 4√3 ≈ 2.31

5 Check

Let's check: put c into f′.

We get 4, the average slope.

Right!

f′(2.3094) ≈ 4

Let's look back.

First: check the conditions.

Then: the average slope.

Then: find c.

Then: keep the ones inside.

Last: check.

Answer