PLEASE HELPPPP
2 answers:
Answer:
C. the range and domain of the graph are the same.
Step-by-step explanation:
A. Is not possible. Here's why
Notice that when the range of a function is all real numbers greater than or equal to 0, then the graph would be above the "x" axis and this graph is below the "x" axis.
B. Is not possible. Here's why
The right side of the "x" axis is not part of the graph at all, therefore the domain can't be all real numbers.
C. This is the right answer.
The graph is below the "x" axis and to the left of the "y" axis, therefore both the domain and the range of the graph are .
D. Is not possible. Here's why
The right side of the "x" axis is not part of the graph, therefore that can't be the answer.
You might be interested in
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.
