Question

Critical points?

Answer 1

Critical points : The point (c, f(c)) is called a critical point of function f(x) if c is in the domain of f(x) and either f ' (c) = 0 or f ' (c) does not exist.

For example the function is f(x) = 4x³ - 27x² - 30x - 4.

Derivative both sides with respected to x.

f ' (x) = 4(3x²) - 27(2x) - 30

f ' (x) = 12x² - 54x - 30.

Again derivative to each side with respect of x.

f " (x) = 12(2x) - 54.

f " (x) = 24x - 54.

To find the critical numbers, to make the first derivative equal to zero or f ' (x) = 0.

12x² - 54x - 30 = 0.

2x² - 9x - 5 = 0.

2x² - 10x + x - 5 = 0.

2x(x - 5) + (x - 5) = 0.

(2x + 1)(x - 5) = 0.

2x + 1 = 0 and x - 5 = 0.

x = - 1/2 and x = 5.

Find Extrema :

To find out extrema, use theorem.

If f " (c) > 0 (positive) ------> minimum point.

If f " (c) < 0 (negative) ------> maximum point.

So, lets plug each critical point in f " (x) = 24x - 54.

If x = - 1/2 then f " (- 1/2) = 24(-1/2) - 54 = - 12 - 54 = - 66 < 0 (negative), therefore maximum point.

If x = 5 then f " (5) = 24(5) - 54 = 120 - 54 = 66 > 0 (positive), therefore minimum point.

To find the f(x) to each x for max and min plugging those values in the original function.

If x = - 1/2 then,

f(-1/2) = 4(-1/2)³ - 27(-1/2)² - 30(-1/2) - 4 = - 4/8 - 27/4 + 15 - 4 =  - 1/2 - 27/4 + 11 = (- 2 - 27 + 44)/4 = 15/4.

If x = 5 then,

f(5) = 4(5)³ - 27(5)² - 30(5) - 4 = 4(125) - 27(25) - 150 - 4 = 500 - 675 -150 - 4 =  - 329.

The relative maximum is f(-1/2) = 15/4 and the relative minimum is f(5) = - 329.

The critical points are (-1/2, 15/4) and (5, -329).