Solve Rolle's theorem for f(x) = -x^2 + 6x on [1, 5] Step by Step

Step by step

The question

Question: find the number c from Rolle's theorem for f(x) = −x² + 6x on [1, 5].

f(x) = −x² + 6x

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, 5], and smooth between.

The theorem can be used.

a polynomial: no breaks, no corners ✓

The ends: f of 1 is 5.

f(1) = −1² + 6(1) = −1 + 6 = 5

And f of 5 is 5.

f(5) = −5² + 6(5) = −25 + 30 = 5

Both are 5.

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

f(1) = f(5) = 5 ✓

2 Find c

The slope: use the power rule.

f′(x) = −2x^(2 − 1) + 6 × 1

Now we have f′(x) = −2x + 6.

= −2x + 6

At c the slope is 0.

So f′(c) = 0.

−2c + 6 = 0

Take away 6 on both sides, so they stay equal.

−2c = −6

Divide by −2: c = 3.

c = −6−2 = 3

3 Keep the ones inside

c = 3 is between 1 and 5.

Keep it.

c = 3 between 1 and 5 ✓

So c = 3.

= c = 3

4 Check

Let's check: put c into f′.

We get 0: flat.

Right!

f′(3) = 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