Use the second derivative test on f(x) = -x^2

Step by step

The question

Question: find the maximum and minimum points of f(x) = −x² − 6x with the second derivative test.

f(x) = −x² − 6x

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

3 Find f″(x)

f double prime is the derivative of f′.

It tells how the slope changes.

f″(x): the derivative of f′(x)

Use the power rule again, now on f′.

f″(x) = −2 × 1 + 0

Now we have f″(x) = −2.

= −2

If f″ is positive, the slope is growing: the curve bends up, like a cup.

Its bottom is a minimum.

If f″ is negative, it bends down, like a cap.

Its top is a maximum.

cup f″ > 0: a cup, a min

cap f″ < 0: a cap, a max

4 Test each point

At x = −3, f″ is −2: negative.

A cap.

So a maximum.

f″(−3) = −2 < 0 cap max

Careful!

Negative f″ is a cap.

The top of a cap is a maximum, not a minimum.

Careful: not f″(−3) < 0 → min ✗

Its y: put −3 into f.

We get 9.

f(−3) = −(−3)² − 6(−3) = −9 + 18 = 9

So: a maximum at (−3, 9).

= max (−3, 9)

5 Check

Let's check with the graph.

A cap at the maximum.

Right!

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: find f″(x).

Then: test each point.

Last: check.

Answer