ANSWER
Yes it is very true
<u>EXPLANATION</u>
If the two equations intersect at then this point must satisfy the two equations.
We substitute in to erquation (1)
We now substitute in to erquation (2) also
Since the point satisfy all the two equations, it is true that they intersect at
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.
