Math Magic: Concavity & Extrema Explorer

Exercise 7.7 - Interactive Learning Journey

Discover the beauty of calculus with colorful step-by-step solutions!

Problem 1(i): Concavity & Inflection Points
f(x) = x(x-4)³
Step 1: Find First Derivative

Using product rule and chain rule:

f'(x) = (x-4)³ + x·3(x-4)²·1
f'(x) = (x-4)² [ (x-4) + 3x ]
f'(x) = (x-4)²(4x-4) = 4(x-4)²(x-1)
Step 2: Find Second Derivative

Differentiate f'(x):

f''(x) = 4[2(x-4)(x-1) + (x-4)²·1]
f''(x) = 4(x-4)[2(x-1) + (x-4)]
f''(x) = 4(x-4)(3x-6) = 12(x-4)(x-2)
Step 3: Find Critical Points

Set f''(x) = 0:

12(x-4)(x-2) = 0
Critical points: x = 2, x = 4
Step 4: Sign Analysis

Test intervals: (-∞, 2), (2, 4), (4, ∞)

f''(x) > 0 when x < 2 → Concave Up
f''(x) < 0 when 2 < x < 4 → Concave Down
f''(x) > 0 when x > 4 → Concave Up

Inflection Points: (2, f(2)) = (2, -8) and (4, f(4)) = (4, 0)

Concave Up: (-∞, 2) ∪ (4, ∞)

Concave Down: (2, 4)

Problem 1(ii): Concavity & Inflection Points
f(x) = sin x + cos x, 0 < x < 2π
Step 1: Find First Derivative
f'(x) = cos x - sin x
Step 2: Find Second Derivative
f''(x) = -sin x - cos x
f''(x) = -(sin x + cos x)
Step 3: Find Critical Points

Set f''(x) = 0:

-(sin x + cos x) = 0
sin x + cos x = 0
tan x = -1
Critical points: x = 3π/4, x = 7π/4
Step 4: Sign Analysis

Test intervals: (0, 3π/4), (3π/4, 7π/4), (7π/4, 2π)

f''(x) < 0 when x ∈ (0, 3π/4) → Concave Down
f''(x) > 0 when x ∈ (3π/4, 7π/4) → Concave Up
f''(x) < 0 when x ∈ (7π/4, 2π) → Concave Down

Inflection Points: (3π/4, 0) and (7π/4, 0)

Concave Up: (3π/4, 7π/4)

Concave Down: (0, 3π/4) ∪ (7π/4, 2π)

Problem 1(iii): Concavity & Inflection Points
f(x) = ½(eˣ - e⁻ˣ)
Step 1: Find First Derivative
f'(x) = ½(eˣ + e⁻ˣ)
Step 2: Find Second Derivative
f''(x) = ½(eˣ - e⁻ˣ)
Step 3: Find Critical Points

Set f''(x) = 0:

½(eˣ - e⁻ˣ) = 0
eˣ = e⁻ˣ
e²ˣ = 1 → x = 0
Step 4: Sign Analysis
f''(x) < 0 when x < 0 → Concave Down
f''(x) > 0 when x > 0 → Concave Up

Inflection Point: (0, 0)

Concave Up: (0, ∞)

Concave Down: (-∞, 0)

Problem 2(i): Local Extrema (2nd Derivative Test)
f(x) = -3x⁵ + 5x³
Step 1: Find First Derivative
f'(x) = -15x⁴ + 15x²
f'(x) = 15x²(1 - x²)
f'(x) = 15x²(1 - x)(1 + x)
Step 2: Find Critical Points

Set f'(x) = 0:

15x²(1 - x)(1 + x) = 0
Critical points: x = -1, x = 0, x = 1
Step 3: Find Second Derivative
f''(x) = -60x³ + 30x
f''(x) = -30x(2x² - 1)
Step 4: Apply 2nd Derivative Test
f''(-1) = -30(-1)(2(1)-1) = 30 > 0 → Local Min
f''(0) = 0 → Test Inconclusive
f''(1) = -30(1)(2-1) = -30 < 0 → Local Max

Local Minimum: at x = -1, f(-1) = -2

Local Maximum: at x = 1, f(1) = 2

At x = 0: Use first derivative test → neither max nor min

Problem 2(ii): Local Extrema (2nd Derivative Test)
f(x) = x ln x
Step 1: Find First Derivative
f'(x) = ln x + x·(1/x)
f'(x) = ln x + 1
Step 2: Find Critical Points

Set f'(x) = 0:

ln x + 1 = 0
ln x = -1
x = e⁻¹ ≈ 0.3679
Step 3: Find Second Derivative
f''(x) = 1/x
Step 4: Apply 2nd Derivative Test
f''(e⁻¹) = 1/(e⁻¹) = e > 0 → Local Minimum

Local Minimum: at x = e⁻¹ ≈ 0.3679, f(e⁻¹) = e⁻¹ ln(e⁻¹) = -e⁻¹ ≈ -0.3679

Problem 2(iii): Local Extrema (2nd Derivative Test)
f(x) = x² e⁻²ˣ
Step 1: Find First Derivative
f'(x) = 2x e⁻²ˣ + x² (-2e⁻²ˣ)
f'(x) = 2e⁻²ˣ (x - x²)
Step 2: Find Critical Points

Set f'(x) = 0:

2e⁻²ˣ (x - x²) = 0
x(1 - x) = 0
Critical points: x = 0, x = 1
Step 3: Find Second Derivative
f''(x) = 2e⁻²ˣ (2x² - 4x + 1)
Step 4: Apply 2nd Derivative Test
f''(0) = 2e⁰ (0 - 0 + 1) = 2 > 0 → Local Min
f''(1) = 2e⁻² (2 - 4 + 1) = -2/e² < 0 → Local Max

Local Minimum: at x = 0, f(0) = 0

Local Maximum: at x = 1, f(1) = e⁻² ≈ 0.1353

Problem 3: Full Analysis
f(x) = 4x³ + 3x² - 6x + 1
Step 1: First Derivative
f'(x) = 12x² + 6x - 6
f'(x) = 6(2x² + x - 1) = 6(2x - 1)(x + 1)
Step 2: Critical Points

Set f'(x) = 0:

6(2x - 1)(x + 1) = 0
Critical points: x = -1, x = 0.5
Step 3: Monotonicity
f'(x) > 0 when x < -1 → Increasing
f'(x) < 0 when -1 < x < 0.5 → Decreasing
f'(x) > 0 when x > 0.5 → Increasing
Step 4: Second Derivative
f''(x) = 24x + 6
f''(x) = 6(4x + 1)
Step 5: Concavity & Inflection

f''(x) = 0 when 4x + 1 = 0 → x = -0.25

f''(x) < 0 when x < -0.25 → Concave Down
f''(x) > 0 when x > -0.25 → Concave Up

Inflection Point: (-0.25, f(-0.25))

Local Max: x = -1, f(-1) = 6

Local Min: x = 0.5, f(0.5) = -1.5

Inflection Point: (-0.25, 2.125)

Key Concepts Explained

Concavity

A function is concave up when its graph curves upward like a U-shape (f''(x) > 0).

A function is concave down when its graph curves downward like an upside-down U (f''(x) < 0).

Inflection points occur where concavity changes - where f''(x) = 0 or undefined.

Local Extrema

Local maximum is the highest point in a small interval (f'(x) = 0 and f''(x) < 0).

Local minimum is the lowest point in a small interval (f'(x) = 0 and f''(x) > 0).

The second derivative test helps determine whether a critical point is max or min.

Monotonicity

A function is increasing when f'(x) > 0 (graph rises as you move right).

A function is decreasing when f'(x) < 0 (graph falls as you move right).

Critical points (where f'(x) = 0) separate increasing/decreasing intervals.