Solve Rolle's theorem for f(x) = x^2 + 2x on [-2, 0] Step by Step

Step by step

The question

Question: find the number c from Rolle's theorem for f(x) = x² + 2x on [−2, 0].

f(x) = x² + 2x

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 [−2, 0], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of −2 is 0.

f(−2) = (−2)² + 2(−2) = 4 − 4 = 0

And f of 0 is 0.

f(0) = 0² + 2(0) = 0 + 0 = 0

Both are 0.

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

f(−2) = f(0) = 0 ✓

2 Find c

The slope: use the power rule.

f′(x) = 2x^(2 − 1) + 2 × 1

Now we have f′(x) = 2x + 2.

= 2x + 2

At c the slope is 0.

So f′(c) = 0.

2c + 2 = 0

Take away 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 −2 and 0.

Keep it.

c = −1 between −2 and 0 ✓

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