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The mean value theorem / Rolle's theorem

Calculus 1 · 22 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Check the conditions
  2. The average slope
  3. Find c
  4. Keep the ones inside
  5. Check

The steps for Mean value theorem for f(x) = 2x^2 - x on [1, 4]. Every solution names its steps, so you always know what comes next.

Worked examples

Solve Rolle's theorem for f(x) = x^2 - 4x on [0, 4] Easy

The question

Question: find the number c from Rolle's theorem for f(x) = x² − 4x on [0, 4].

f(x) = x2 − 4x

The theorem: if f is the same at both ends, then for some c between them, the slope is 0.

f(a) = f(b) → f′(c) = 0
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) = 02 − 4(0) = 0 + 0 = 0

And f of 4 is 0.

f(4) = 42 − 4(4) = 16 − 16 = 0

Both are 0.

The same height, so Rolle's theorem works here.

f(0) = f(4) = 0 ✓

2 Find c

The slope: use the power rule.

f′(x) = 2x^(2 − 1) − 4 × 1

Now we have f′(x) = 2x − 4.

= 2x − 4

At c the slope is 0.

So f′(c) = 0.

2c − 4 = 0

Add 4 on both sides, so they stay equal.

2c = 4

Divide by 2: c = 2.

c = 42 = 2

3 Keep the ones inside

c = 2 is between 0 and 4.

Keep it.

c = 2 between 0 and 4 ✓

So c = 2.

= c = 2

4 Check

Let's check: put c into f′.

We get 0: flat.

Right!

f′(2) = 0

And in a picture.

The two ends are at the same height.

At c, the curve is flat: its tangent line is level.

Right!

Let's look back.

First: check the conditions.

Then: find c.

Then: keep the ones inside.

Last: check.

Answer
= c = 2

▶ Watch it solved

Mean value theorem for f(x) = 2x^2 - x on [1, 4] Medium

The question

Question: find the number c from the mean value theorem for f(x) = 2x² − x on [1, 4].

f(x) = 2x2 − x

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, 4], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of 1 is 1.

f(1) = 2(1)2 − 1 = 2 − 1 = 1

And f of 4 is 28.

f(4) = 2(4)2 − 4 = 32 − 4 = 28

2 The average slope

The average slope from 1 to 4: the change in f over the change in x.

It is 9.

f(4) − f(1)4 − 1 = 28 − 14 − 1 = 273 = 9

Careful!

Divide by the width, 4 minus 1.

The average slope is 9, not 27.

Careful: not f(4) − f(1) = 27 ✗

3 Find c

The slope: use the power rule.

f′(x) = 2 × 2x^(2 − 1) − 1

Now we have f′(x) = 4x − 1.

= 4x − 1

At c the slope is the average slope: f′(c) = 9.

4c − 1 = 9

Add 1 on both sides, so they stay equal.

4c = 10

Divide by 4: c = 5/2.

c = 104 = 52

4 Keep the ones inside

c = 5/2 is between 1 and 4.

Keep it.

c = 52 ≈ 2.5 between 1 and 4 ✓

So c = 5/2.

= c = 52 ≈ 2.5

5 Check

Let's check: put c into f′.

We get 9, the average slope.

Right!

f′(52) = 9

Let's look back.

First: check the conditions.

Then: the average slope.

Then: find c.

Then: keep the ones inside.

Last: check.

Answer
= c = 52 ≈ 2.5

▶ Watch it solved

Find c for the mean value theorem, f(x) = x^3 - 3x on [-2, 3] Harder

The question

Question: find the number c from the mean value theorem for f(x) = x³ − 3x on [−2, 3].

f(x) = x3 − 3x

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 [−2, 3], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of −2 is −2.

f(−2) = (−2)3 − 3(−2) = −8 + 6 = −2

And f of 3 is 18.

f(3) = 33 − 3(3) = 27 − 9 = 18

2 The average slope

The average slope from −2 to 3: the change in f over the change in x.

It is 4.

f(3) − f(−2)3 − −2 = 18 − (−2)3 − (−2) = 205 = 4

Careful!

Divide by the width, 3 minus −2.

The average slope is 4, not 20.

Careful: not f(3) − f(−2) = 20 ✗

3 Find c

The slope: use the power rule.

f′(x) = 3x^(3 − 1) − 3 × 1

Now we have f′(x) = 3x² − 3.

= 3x2 − 3

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

3c2 − 3 = 4

Add 3 on both sides.

3c2 = 7

Divide by 3.

c2 = 73

Take the square root.

c is plus or minus √7 over √3, about 1.53.

c = √7√3 or −√7√3

4 Keep the ones inside

c, about 1.53, is between −2 and 3.

Keep it.

c = √7√3 ≈ 1.53 between −2 and 3 ✓

c, about −1.53, is between −2 and 3.

Keep it.

c = −√7√3 ≈ −1.53 between −2 and 3 ✓

So c = √7 over √3, about 1.53, and c = −√7 over √3, about −1.53.

= c = √7√3 ≈ 1.53 and c = −√7√3 ≈ −1.53

5 Check

Let's check: put c into f′.

We get 4, the average slope.

Right!

f′(1.5275) ≈ 4

Let's look back.

First: check the conditions.

Then: the average slope.

Then: find c.

Then: keep the ones inside.

Last: check.

Answer
= c = √7√3 ≈ 1.53 and c = −√7√3 ≈ −1.53

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. Rolle's theorem for f(x) = x^2 - 4x on [0, 4]
    Answer
    = c = 2
  2. Rolle's theorem for f(x) = x^2 - 2x - 3 on [-1, 3]
    Answer
    = c = 1
  3. Rolle's theorem for f(x) = x^2 + 2x on [-2, 0]
    Answer
    = c = −1
  4. Rolle's theorem for f(x) = -x^2 + 6x on [1, 5]
    Answer
    = c = 3
  5. mean value theorem for f(x) = x^2 on [1, 3]
    Answer
    = c = 2
  6. mean value theorem for f(x) = x^2 - 2x on [0, 3]
    Answer
    = c = 32 ≈ 1.5
  7. mean value theorem for f(x) = x^2 + 4x on [0, 2]
    Answer
    = c = 1

Medium

  1. mean value theorem for f(x) = 2x^2 - x on [1, 4]
    Answer
    = c = 52 ≈ 2.5
  2. mean value theorem for f(x) = x^2 - 6x + 1 on [0, 5]
    Answer
    = c = 52 ≈ 2.5
  3. mean value theorem for f(x) = -x^2 + 3x on [-1, 2]
    Answer
    = c = 12 ≈ 0.5
  4. mean value theorem for f(x) = 3x^2 + 1 on [0, 2]
    Answer
    = c = 1
  5. find c for the mean value theorem, f(x) = x^3 - x on [0, 2]
    Answer
    = c = 2√3 ≈ 1.15
  6. find c for the mean value theorem, f(x) = x^3 on [0, 3]
    Answer
    = c = √3 ≈ 1.73
  7. mean value theorem for f(x) = x^3 - 2x on [0, 2]
    Answer
    = c = 2√3 ≈ 1.15

Harder

  1. find c for the mean value theorem, f(x) = x^3 - 3x on [-2, 3]
    Answer
    = c = √7√3 ≈ 1.53 and c = −√7√3 ≈ −1.53
  2. find c for the mean value theorem, f(x) = x^3 + x on [-1, 2]
    Answer
    = c = 1
  3. mean value theorem for f(x) = x^3 - 12x on [0, 4]
    Answer
    = c = 4√3 ≈ 2.31
  4. mean value theorem for f(x) = x^3 + 2x on [-2, 2]
    Answer
    = c = 2√3 ≈ 1.15 and c = −2√3 ≈ −1.15
  5. Rolle's theorem for f(x) = x^3 - 4x on [-2, 2]
    Answer
    = c = 2√3 ≈ 1.15 and c = −2√3 ≈ −1.15
  6. Rolle's theorem for f(x) = x^3 - x on [0, 1]
    Answer
    = c = 1√3 ≈ 0.58
  7. mean value theorem for f(x) = x^3 on [-1, 2]
    Answer
    = c = 1
  8. mean value theorem for f(x) = x^3 - 3x^2 on [0, 3]
    Answer
    = c = 2