Solve Rolle's theorem for f(x) = x^2 - 2x - 3 on [-1, 3] Step by Step

Step by step

The question

Question: find the number c from Rolle's theorem for f(x) = x² − 2x − 3 on [−1, 3].

f(x) = x² − 2x − 3

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 [−1, 3], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of −1 is 0.

f(−1) = (−1)² − 2(−1) − 3 = 1 + 2 − 3 = 0

And f of 3 is 0.

f(3) = 3² − 2(3) − 3 = 9 − 6 − 3 = 0

Both are 0.

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

f(−1) = f(3) = 0 ✓

2 Find c

The slope: use the power rule.

f′(x) = 2x^(2 − 1) − 2 × 1 + 0

Now we have f′(x) = 2x − 2.

= 2x − 2

At c the slope is 0.

So f′(c) = 0.

2c − 2 = 0

Add 2 on both sides, so they stay equal.

2c = 2

Divide by 2: c = 1.

c = 22 = 1

3 Keep the ones inside

c = 1 is between −1 and 3.

Keep it.

c = 1 between −1 and 3 ✓

So c = 1.

= c = 1

4 Check

Let's check: put c into f′.

We get 0: flat.

Right!

f′(1) = 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