A function f(x) has range {yer 1-50}. Explain why reflecting the function vertically would not affect the range

For the reflection property, we have;

For reflection vertically, which is equivalent to reflection across the y-axis, we have;

Let the location or point of the pre-image before, reflection be (x, y), then, the location or point of the image after reflection will be (-x, y)

Therefore, the y-coordinates remains the same

Given that the range of the function is the value of f(x) or y from 1 - 50, which are the y-coordinate values of the function, it will not be affected by the vertical reflection of the function.

Step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}