Answer:
y = x/2 - 4 slope-intercept form
y = x/2 + 4 slope-intercept form
NO SOLUTION.
The graph is a pair of parallel lines one line through point (0, -4) and the other through point (0, 4)
Step-by-step explanation:
x - 2y = 8 and -2x+4y=-16.
we can easily convert these to slope intercept form
x - 2y = 8
-2y = 8 - x
2y = x - 8
y = x/2 - 8/2
y = x/2 - 4 slope-intercept form
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-2x + 4y = 16
4y = 2x + 16
y = 2x/4 + 16/4
y = x/2 + 4 slope-intercept form
We have 2 lines that are parallel because they have the same slope
These piecewise functions are several graphs pieced together. Where one ends the other begins.
The greater than (> or <) symbol means an open circle at that point. If it is greater than or equal to (>=) then you put a closed solid circle.
start with the first one
-2x + 3 where -4 =< x < -1
put in these values for x
solid circle at (-4, 11)
open circle at (-1, 5)
graph is a straight line so connect the circles
next one
|2x -1| where -1 < x < 2
absolute value graphs are V-SHAPED where y is always positive. solve for vertex (x, 0)
0 = 2x -1
x = 1/2
(1/2, 0) vertex of V
solve for endpoints, both are open circles
(-1, 3)
(2, 3)
next one is easy. it's just a solid point
When y = 5 ; x = 2
(2,5)
last is a parabola, or just half of a parabola, aka bowl shaped graph
3x^2 - 1 when x > 2
plug in x = 2
open circle at (2, 11)
you can put in another x value greater than 2 to find a second point to help draw the graph but it's just a curve upward that continues with an arrow, no endpoint
Answer:
The correct option is;
D) Neither the domain restriction nor the expression for f⁻¹(x) is correct
Step-by-step explanation:
The given function can be written as follows;
f(x) = x² - 331
The inverse of the function f(x), which is f⁻¹(x), can be found as follows;
Let f(x) = x² - 331 = y, we have;
f(x) + 331 = x²
x = √(f(x) + 331)
We replace x with f⁻¹(x) and f(x) with x, to get;
f⁻¹(x) = √(x + 331)
We note that the domain of the inverse function will include values from x ≥ -331, and the correct inverse function √(x + 331) ≠ √x - 331
Therefore, we have that neither the domain restriction nor the expression for f⁻¹(x) is correct.
