What is the value of x in the equation 3(2x + 8) = 0? (4 points)
1 answer:
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Answer:
(-∞,-3) and (3,∞)
Step-by-step explanation:
f(x) = x³ − 27x + 3
1. Find the critical points
(a) Calculate the first derivative of the function.
f'(x) = 3x² -27
(b) Factor the first derivative
f'(x)= 3(x² - 9) = 3(x + 3) (x - 3)
(c) Find the zeros
3(x + 3) (x - 3) = 0
x + 3 = 0 x - 3 = 0
x = -3 x = 3
The critical points are at <em>x = -3</em> and x = 3.
2. Find the local extrema
(a) x = -3
f(x) = x³ − 27x + 3 = (-3)³ - 27(-3) + 3 = -27 +81 + 3 = 57
(b) x = 3
f(x) = x³ − 27x + 3 = 3³ - 27(3) + 3 = 27 - 81 + 3 = -51
The local extrema are at (-3,57) and (3,-51).
3, Identify the local extrema as maxima or minima
Test the first derivative (the slope) over the intervals (-∞, -3), (-3,3), (3,∞)
f'(-4) = 3x² -27 = 3(4)² - 27 = 21
f'(0) = 3(0)² -27 = -27
f'(4) = 3(4)² - 27 = 51
The function is increasing on the intervals (-∞,-3) and (3,∞).
The graph below shows the critical points of your function.
