Mean value theorem for f(x) = x^2 - 6x + 1 on [0, 5] Step by Step

Step by step

The question

Question: find the number c from the mean value theorem for f(x) = x² − 6x + 1 on [0, 5].

f(x) = x² − 6x + 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, 5], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of 0 is 1.

f(0) = 0² − 6(0) + 1 = 0 + 0 + 1 = 1

And f of 5 is −4.

f(5) = 5² − 6(5) + 1 = 25 − 30 + 1 = −4

2 The average slope

The average slope from 0 to 5: the change in f over the change in x.

It is −1.

f(5) − f(0)5 − 0 = −4 − 15 − 0 = −55 = −1

Careful!

Divide by the width, 5 minus 0.

The average slope is −1, not −5.

Careful: not f(5) − f(0) = −5 ✗

3 Find c

The slope: use the power rule.

f′(x) = 2x^(2 − 1) − 6 × 1 + 0

Now we have f′(x) = 2x − 6.

= 2x − 6

At c the slope is the average slope: f′(c) = −1.

2c − 6 = −1

Add 6 on both sides, so they stay equal.

2c = 5

Divide by 2: c = 5/2.

c = 52 = 52

4 Keep the ones inside

c = 5/2 is between 0 and 5.

Keep it.

c = 52 ≈ 2.5 between 0 and 5 ✓

So c = 5/2.

= c = 52 ≈ 2.5

5 Check

Let's check: put c into f′.

We get −1, the average slope.

Right!

f′(52) = −1

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