The graph of f (x) = −x3 + 3x2 − 2 is shown in the figure to the right.
Complete the table below by using
the second derivative of f to identify
the intervals on which f us concave
upward or concave downward and
to identify the inflection points of f.

Concave Up: (-∞, 1)

Concave Down: (1, ∞)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDASEquality Properties

Algebra I

Combining Like Terms

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Second Derivative Test:

Possible Points of Inflection (P.P.I) - Tells us the possible x-values where the graph f(x) may change concavity. Occurs when f"(x) = 0 or undefinedPoints of Inflection (P.I) - Actual x-values when the graph f(x) changes concavityNumber Line Test - Helps us determine whether a P.P.I is a P.I

Step-by-step explanation:

Step 1: Define

f(x) = -x³ + 3x² - 2

Step 2: Find 2nd Derivative

1st Derivative [Basic Power]:                    f'(x) = -1 · 3x³⁻² + 2 · 3x²⁻¹Simplify:                                                     f'(x) = -3x² + 6x2nd Derivative [Basic Power]:                  f"(x) = -3 · 2x²⁻¹ + 1 · 6x¹⁻¹Simplify:                                                     f"(x) = -6x + 6

Step 3: Find P.P.I

Set f"(x) equal to zero:                    0 = -6x + 6Isolate x term:                                 -6 = -6xIsolate x:                                          1 = xRewrite:                                           P.P.I x = 1

Step 4: Number Line Test

See Attachment.

We plug in the test points into the 2nd Derivative and see if the P.P.I is a P.I.

x = 0

Substitute:                    f"(0) = -6(0) + 6Multiply:                        f"(0) = 0 + 6Add:                              f"(0) = 6

This means that the graph f(x) is concave up before x = 1.

x = 2

Substitute:                    f"(2) = -6(2) + 6Multiply:                        f"(2) = -12 + 6Add:                              f"(2) = -6

This means that the graph f(x) is concave down after x = 1.

Step 5: Identify

Since f"(x) changes concavity from positive to negative at x = 1, then we know that the P.P.I x = 1 is actually a P.I x = 1.

Let's find what actual point on f(x) when the concavity changes.

Substitute in P.I into f(x):                    f(1) = -(1)³ + 3(1)² - 2Evaluate Exponents:                          f(1) = -(1) + 3(1) - 2Multiply:                                              f(1) = -1 + 3 - 2Combine like terms:                          f(1) = 0

So at (1, 0), f(x) changes concavity from concave up to concave down.

Step 6: Define Intervals

We know that before f(x) reaches x = 1, the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval: (-∞, 1)

We also know that after f(x) passes x = 1, the graph is concave down. We used the 2nd Derivative Test to confirm this as well.

Concave Down Interval: (1, ∞)