Use the second derivative test on f(x) = -x^3 + 2 Step by Step

Step by step

The question

Question: find the maximum and minimum points of f(x) = −x³ + 2 with the second derivative test.

f(x) = −x³ + 2

1 Find f′(x)

Use the power rule on each term.

f′(x) = −3x^(3 − 1) + 0

Now we have f′(x) = −3x².

= −3x²

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

−3x² = 0

−3x² is 0 only when x is 0.

−3x² = 0 → x = 0

So x = 0.

x = 0

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) = −3 × 2x^(2 − 1)

Now we have f″(x) = −6x.

= −6x

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 = 0, f″ is 0.

The test cannot tell.

f″(0) = −6(0) = 0

So we look at the slope on each side, as in the max and min lesson.

Down before, down after.

It does not turn: not a max, not a min.

f′(−1) = −3 < 0 ↘, f′(1) = −3 < 0 ↘

So there is no maximum or minimum point.

= no maximum or minimum

5 Check

Let's check with the graph.

It never turns.

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