Help me please?

Question

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!

Answer 1

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

Answer 2

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)

Answer 3

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.



====

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)