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

Step by step

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