For f(x) = x^3 − 27x: Find all critical points of f(x), and classify each one as either a relative.
My work is attached, can you guys help me figure out where I am going wrong? Submitted the answer to my instructor and he said there's a careless error.
Thank you!
Help me please?
回答 (3)
2018-06-04 3:40 pm
✔ 最佳答案
f(x) = x^3 - 27x.First derivative: f'(x) = 3x^2 - 27 (✔).
Critical (or stationary) points: x = ±9 (❌).
f'(x) = 0 ===> 3x^2 - 27 = 0 ===> 3x^2 = 27 ===> x^2 = 27/3 = 9 ===> x = ±√9 = ±3 (✔).
(*) When: x = -3, f(-3) = (-3)^3 - 27(-3) = -27 + 81 = 54 ===> (-3; 54) is a critical point.
(**) When: x = 3, f(3) = 3^3 - 27(3) = 27 - 81 = -54 ===> (3; -54) is a critical point.
Second derivative test:
f''(x) = [f'(x)]' = (3x^2 - 27)' = 6x.
When: x = -3 ===> f''(-3) = 6(-3) = -18 (< 0) ===> (-3; 54) is a (relative) maximum point.
When: x = 3 ===> f''(3) = 6(3) = 18 (> 0) ===> (3; -54) is a (relative) minimum point.
Graph: https://www.desmos.com/calculator/nlkmzzsbao
2018-06-04 3:53 pm
In your answer :
Firstly, 3x² - 27 = 0
will lead to x = ±3 but x ≠ ±9
Secondly, failed to use f'(x) to determine maximum and minimum.
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My answer:
f(x) = x³ - 27x
f'(x) = 3x² - 27 = 3(x + 3)(x - 3)
f"(x) = 6x
When x = -3:
f'(x) = 0
f(x) = (-3)³ - 27(-3) = 54
f"(x) = 6(-3) = -18 < 0
Hence, maximum at (-3, 54)
When x = 3:
f'(x) = 0
f(x) = (3)³ - 27(3) = -54
f"(x) = 6(3) = 18 > 0
Hence, minimum at (3, -54)
Firstly, 3x² - 27 = 0
will lead to x = ±3 but x ≠ ±9
Secondly, failed to use f'(x) to determine maximum and minimum.
====
My answer:
f(x) = x³ - 27x
f'(x) = 3x² - 27 = 3(x + 3)(x - 3)
f"(x) = 6x
When x = -3:
f'(x) = 0
f(x) = (-3)³ - 27(-3) = 54
f"(x) = 6(-3) = -18 < 0
Hence, maximum at (-3, 54)
When x = 3:
f'(x) = 0
f(x) = (3)³ - 27(3) = -54
f"(x) = 6(3) = 18 > 0
Hence, minimum at (3, -54)
2018-06-04 7:31 pm
You can obtain the critical points when the derivative is null.
f(x) = x³ - 27x ← this is a curve, i.e. a function
f’(x) = 3x² - 27 ← this is the derivative, then you solve for x,: f’(x) = 0
3x² - 27 = 0
3x² = 27
x² = 9
x = ± 3
f(x) = x³ - 27x → when: x = 3
f(3) = 27 - 81 = - 54
→ First point (3 ; - 54)
f(x) = x³ - 27x → when: x = - 3
f(3) = - 27 + 81 = 54
→ Second point (- 3 ; 54)
f(x) = x³ - 27x ← this is a curve, i.e. a function
f’(x) = 3x² - 27 ← this is the derivative, then you solve for x,: f’(x) = 0
3x² - 27 = 0
3x² = 27
x² = 9
x = ± 3
f(x) = x³ - 27x → when: x = 3
f(3) = 27 - 81 = - 54
→ First point (3 ; - 54)
f(x) = x³ - 27x → when: x = - 3
f(3) = - 27 + 81 = 54
→ Second point (- 3 ; 54)
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