Exercise 7.6 - Absolute Extrema
Find the absolute maximum and minimum values of functions on given intervals
(i) \( f(x) = x^2 - 12x + 10 \) on \([1, 7]\)
Show Solution
1
Find the derivative:
f'(x) = 2x - 12
2
Find critical points:
2x - 12 = 0 ⇒ x = 6
Critical point at x = 6 .
3
Evaluate function at critical point and endpoints:
f(1) = (1)² - 12(1) + 10 = -1
f(6) = (6)² - 12(6) + 10 = -26
f(7) = (7)² - 12(7) + 10 = -25
4
Determine absolute extrema:
Absolute maximum: -1 at x = 1
Absolute minimum: -26 at x = 6
Solution: Absolute max = -1 at x=1, Absolute min = -26 at x=6
Did You Know?
For quadratic functions, the vertex (which gives the minimum or maximum) is always at x = -b/(2a). Here, -b/(2a) = 12/2 = 6, which matches our critical point!
(ii) \( f(x) = 3x^4 - 4x^3 \) on \([-1, 2]\)
Show Solution
1
Find the derivative:
f'(x) = 12x³ - 12x²
2
Find critical points:
12x³ - 12x² = 0 ⇒ 12x²(x - 1) = 0 ⇒ x = 0 or x = 1
Critical points at x = 0 and x = 1 .
3
Evaluate function at critical points and endpoints:
f(-1) = 3(-1)⁴ - 4(-1)³ = 3 + 4 = 7
f(0) = 0 - 0 = 0
f(1) = 3(1)⁴ - 4(1)³ = 3 - 4 = -1
f(2) = 3(16) - 4(8) = 48 - 32 = 16
4
Determine absolute extrema:
Absolute maximum: 16 at x = 2
Absolute minimum: -1 at x = 1
Solution: Absolute max = 16 at x=2, Absolute min = -1 at x=1
(iii) \( f(x) = 6x^{\frac{4}{3}} - 3x^{\frac{1}{3}} \) on \([-1, 1]\)
Show Solution
1
Find the derivative:
f'(x) = 8x^{1/3} - x^{-2/3} = (8x - 1)/x^{2/3}
2
Find critical points:
f'(x) = 0 ⇒ 8x - 1 = 0 ⇒ x = 1/8
f'(x) undefined at x = 0
Critical points at x = 0 and x = 1/8 .
3
Evaluate function at critical points and endpoints:
f(-1) = 6(-1)^{4/3} - 3(-1)^{1/3} = 6 + 3 = 9
f(0) = 0 - 0 = 0
f(1/8) ≈ 6(0.125)^{1.333} - 3(0.125)^{0.333} ≈ -0.472
f(1) = 6 - 3 = 3
4
Determine absolute extrema:
Absolute maximum: 9 at x = -1
Absolute minimum: -0.472 at x = 1/8
Solution: Absolute max = 9 at x=-1, Absolute min ≈ -0.472 at x=1/8
(iv) \( f(x) = 2\cos x + \sin 2x \) on \(\left[0, \frac{\pi}{2}\right]\)
Show Solution
1
Find the derivative:
f'(x) = -2sinx + 2cos2x
2
Find critical points:
-2sinx + 2cos2x = 0 ⇒ cos2x = sinx
Using identity: 1 - 2sin²x = sinx ⇒ 2sin²x + sinx - 1 = 0
Solutions: sinx = 0.5 ⇒ x = π/6 (30°)
Critical point at x = π/6 .
3
Evaluate function at critical point and endpoints:
f(0) = 2cos0 + sin0 = 2 + 0 = 2
f(π/6) = 2cos(π/6) + sin(π/3) ≈ 1.732 + 0.866 ≈ 2.598
f(π/2) = 2cos(π/2) + sin(π) = 0 + 0 = 0
4
Determine absolute extrema:
Absolute maximum: ≈2.598 at x = π/6
Absolute minimum: 0 at x = π/2
Solution: Absolute max ≈ 2.598 at x=π/6, Absolute min = 0 at x=π/2
Exercise 7.6 - Monotonicities and Local Extrema
Find intervals where functions increase/decrease and identify local maxima/minima
(i) \( f(x) = 2x^3 + 3x^2 - 12x \)
Show Solution
1
Find the derivative:
f'(x) = 6x² + 6x - 12
2
Find critical points:
6x² + 6x - 12 = 0 ⇒ x² + x - 2 = 0 ⇒ (x+2)(x-1) = 0
Critical points at x = -2 and x = 1 .
3
Test intervals around critical points:
Choose test points: x = -3, x = 0, x = 2
f'(-3) = 6(9) + 6(-3) - 12 = 54 - 18 - 12 = 24 > 0
f'(0) = 0 + 0 - 12 = -12 < 0
f'(2) = 6(4) + 6(2) - 12 = 24 + 12 - 12 = 24 > 0
4
Determine monotonicity:
Increasing on: (-∞, -2) and (1, ∞)
Decreasing on: (-2, 1)
5
Identify local extrema:
Local maximum at x = -2 (changes from + to -)
Local minimum at x = 1 (changes from - to +)
Solution: Increasing on (-∞,-2)∪(1,∞), Decreasing on (-2,1)
(ii) \( f(x) = \frac{x}{x-5} \)
Show Solution
1
Find the derivative (using quotient rule):
f'(x) = [(1)(x-5) - (x)(1)]/(x-5)² = -5/(x-5)²
2
Find critical points:
f'(x) = 0 ⇒ -5/(x-5)² = 0 ⇒ No solution
f'(x) undefined at x = 5 (not in domain)
No critical points.
3
Analyze derivative sign:
(x-5)² is always positive ⇒ f'(x) = -5/(x-5)² is always negative
4
Determine monotonicity:
Decreasing on: (-∞,5) and (5,∞)
Never increasing
5
Identify local extrema:
No local maxima or minima
Solution: Always decreasing on its domain (-∞,5)∪(5,∞)
(iii) \( f(x) = \frac{e^x}{1-e^x} \)
Show Solution
1
Find the derivative (using quotient rule):
f'(x) = [e^x(1-e^x) - e^x(-e^x)]/(1-e^x)² = e^x/(1-e^x)²
2
Find critical points:
f'(x) = 0 ⇒ e^x = 0 ⇒ No solution (e^x always positive)
f'(x) undefined when 1-e^x = 0 ⇒ x = 0
No critical points, vertical asymptote at x=0
3
Analyze derivative sign:
Numerator (e^x) always positive
Denominator (1-e^x)² always positive
f'(x) > 0 for all x ≠ 0
4
Determine monotonicity:
Increasing on: (-∞,0) and (0,∞)
5
Identify local extrema:
No local maxima or minima
Solution: Always increasing on its domain (-∞,0)∪(0,∞)
(iv) \( f(x) = \frac{x^3}{3} - \log x \)
Show Solution
1
Find the derivative:
f'(x) = x² - 1/x
2
Find critical points:
x² - 1/x = 0 ⇒ x³ - 1 = 0 ⇒ x = 1
Critical point at x = 1 .
3
Test intervals around critical point (domain x > 0):
Choose test points: x = 0.5, x = 2
f'(0.5) = 0.25 - 2 = -1.75 < 0
f'(2) = 4 - 0.5 = 3.5 > 0
4
Determine monotonicity:
Decreasing on: (0,1)
Increasing on: (1,∞)
5
Identify local extrema:
Local minimum at x = 1 (changes from - to +)
Solution: Decreasing on (0,1), Increasing on (1,∞)
(v) \( f(x) = \sin x \cos x + 5, \, x \in (0, 2\pi) \)
Show Solution
1
Simplify using identity:
sinx cosx = (1/2)sin2x ⇒ f(x) = (1/2)sin2x + 5
2
Find the derivative:
f'(x) = cos2x
3
Find critical points:
cos2x = 0 ⇒ 2x = π/2, 3π/2, 5π/2, 7π/2 ⇒ x = π/4, 3π/4, 5π/4, 7π/4
4
Test intervals around critical points:
Test points: x = 0, π/2, π, 3π/2, 2π
f'(0) = cos0 = 1 > 0
f'(π/2) = cosπ = -1 < 0
f'(π) = cos2π = 1 > 0
f'(3π/2) = cos3π = -1 < 0
5
Determine monotonicity:
Increasing on: (0,π/4) , (3π/4,5π/4) , (7π/4,2π)
Decreasing on: (π/4,3π/4) , (5π/4,7π/4)
6
Identify local extrema:
Local maxima at x = π/4 and x = 5π/4
Local minima at x = 3π/4 and x = 7π/4
Solution: Increasing on (0,π/4)∪(3π/4,5π/4)∪(7π/4,2π)