Turning points of f(x) = -x^2 + 6x Step by Step

Step by step

The question

Question: find the maximum and minimum points of f(x) = −x² + 6x.

f(x) = −x² + 6x

The derivative is the slope.

Where it is positive, the graph goes up.

Where it is negative, the graph goes down.

f′(x) > 0: going up ↗

f′(x) < 0: going down ↘

1 Find f′(x)

Use the power rule on each term.

f′(x) = −2x^(2 − 1) + 6 × 1

Now we have f′(x) = −2x + 6.

= −2x + 6

2 Solve f′(x) = 0

At a turning point the graph is flat for a moment: the slope is 0.

So we solve f′(x) = 0.

f′(x) = 0

−2x + 6 = 0

Take out the −2.

−2(x − 3) = 0

So x = 3.

x = 3

Careful!

f′ = 0 alone does not tell max or min.

The signs on each side tell us.

Careful: f′(x) = 0 is not enough: look at the signs

3 Sign table

A sign table.

The numbers 3 cut the line into 2 parts.

Test one number in each part: 2, 4.

f′(2) = 2, f′(4) = −2

f′(2) is 2: positive.

So f goes up there.

f′(4) is −2: negative.

So f goes down there.

4 Max and min

At x = 3, f goes up, then down.

That is a maximum.

max

Its y: put 3 into f.

We get 9.

f(3) = −3² + 6(3) = −9 + 18 = 9

So: a maximum at (3, 9).

= max (3, 9)

5 Check

Let's check with the graph.

Up, then down: just like the table.

Right!

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: sign table.

Then: max and min.

Last: check.

Answer