40 pts ! what is the range of the function graphed below?

Explanation:

The range is the set of possible y outputs of a function.

What's the smallest y can get? The answer here is there is no smallest y value as the graph heads off downward forever. So we say negative infinity is the lower bound. So we can say -infinity < y

What's the largest y can get? That would be y = 2. It can't actually get to y = 2 as this is a horizontal asymptote. The graph approaches this horizontal line getting closer and closer, but never actually arriving there. So we can say y < 2

Combine -infinity < y and y < 2 to get -infinity < y < 2

It's much simpler to write y < 2 without the "-infinity < y" portion, but the more expanded out way helps us convert to interval notation which would be (-infinity, 2). The curved parenthesis tell the reader "do not include the endpoint as part of the set". Infinity is not a number, and you can't reach infinity, so there is no way to include it as a set of real numbers.

step-by-step explanation:

if x = 2 you have to find x to finish the problem

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}