Solve Rolle's theorem for f(x) = x^3 - x on [0, 1] Step by Step

Step by step

The question

Question: find the number c from Rolle's theorem for f(x) = x³ − x on [0, 1].

f(x) = x³ − x

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, 1], 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 = 0 + 0 = 0

And f of 1 is 0.

f(1) = 1³ − 1 = 1 − 1 = 0

Both are 0.

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

f(0) = f(1) = 0 ✓

2 Find c

The slope: use the power rule.

f′(x) = 3x^(3 − 1) − 1

Now we have f′(x) = 3x² − 1.

= 3x² − 1

At c the slope is 0.

So f′(c) = 0.

3c² − 1 = 0

Add 1 on both sides.

3c² = 1

Divide by 3.

c² = 13

Take the square root.

c is plus or minus 1 over √3, about 0.58.

c = 1√3 or −1√3

3 Keep the ones inside

c, about 0.58, is between 0 and 1.

Keep it.

c = 1√3 ≈ 0.58 between 0 and 1 ✓

c, about −0.58, is not between 0 and 1.

It does not count.

c = −1√3 ≈ −0.58 not between 0 and 1 ✗

Careful!

The theorem only talks about c between a and b.

Throw away the others.

Careful: c must be between 0 and 1

So c = 1 over √3, about 0.58.

= c = 1√3 ≈ 0.58

4 Check

Let's check: put c into f′.

We get 0: flat.

Right!

f′(0.5774) ≈ 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