contestada

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.

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 class=

Respuesta :

Space

Answer:

Concave Up: (-∞, 1)

Concave Down: (1, ∞)

General Formulas and Concepts:

Pre-Algebra

  • Order of Operations: BPEMDAS
  • Equality Properties

Algebra I

  • Combining Like Terms

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(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 undefined
  • Points of Inflection (P.I) - Actual x-values when the graph f(x) changes concavity
  • Number 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

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

Step 3: Find P.P.I

  1. Set f"(x) equal to zero:                    0 = -6x + 6
  2. Isolate x term:                                 -6 = -6x
  3. Isolate x:                                          1 = x
  4. Rewrite:                                           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

  1. Substitute:                    f"(0) = -6(0) + 6
  2. Multiply:                        f"(0) = 0 + 6
  3. Add:                              f"(0) = 6

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

x = 2

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

  1. Substitute in P.I into f(x):                    f(1) = -(1)³ + 3(1)² - 2
  2. Evaluate Exponents:                          f(1) = -(1) + 3(1) - 2
  3. Multiply:                                              f(1) = -1 + 3 - 2
  4. Combine 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, ∞)

Ver imagen Space

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