Mean value theorem for f(x) = 3x^2 + 1 on [0, 2] Step by Step

Step by step

The question

Question: find the number c from the mean value theorem for f(x) = 3x² + 1 on [0, 2].

f(x) = 3x² + 1

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 1.

f(0) = 3(0)² + 1 = 0 + 1 = 1

And f of 2 is 13.

f(2) = 3(2)² + 1 = 12 + 1 = 13

2 The average slope

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

It is 6.

f(2) − f(0)2 − 0 = 13 − 12 − 0 = 122 = 6

Careful!

Divide by the width, 2 minus 0.

The average slope is 6, not 12.

Careful: not f(2) − f(0) = 12 ✗

3 Find c

The slope: use the power rule.

f′(x) = 3 × 2x^(2 − 1) + 0

Now we have f′(x) = 6x.

= 6x

At c the slope is the average slope: f′(c) = 6.

6c = 6

Divide by 6: c = 1.

c = 66 = 1

4 Keep the ones inside

c = 1 is between 0 and 2.

Keep it.

c = 1 between 0 and 2 ✓

So c = 1.

= c = 1

5 Check

Let's check: put c into f′.

We get 6, the average slope.

Right!

f′(1) = 6

Let's look back.

First: check the conditions.

Then: the average slope.

Then: find c.

Then: keep the ones inside.

Last: check.

Answer