Given the graph of the function f(x) below, what happens to f(x) when x is a very small positive number? F(x) is a negative number with a large absolute value
F(x) is a negatice number with a small absolute value
F(x) is a very small positive number
F(x) is a very large positive number

A) F(x) is a negative number with a large absolute value

Step-by-step explanation:

You have to keep in mind the end behavior of a function like this:

As y -> -infinity, x -> infinity

As y -> infinity, x -> -infinity

Simply put:

As y gets smaller, x gets larger

As y gets larger, x gets smaller

Since the question asks what happens to f(x) (which is y) when x is a very small positive number, y is going to get smaller, but the limit is the vertical asymptote x=0 since that's the smallest positive number (but 0 is not included since it's neither positive or negative). So as x approaches 0, f(x) is going to approach -infinity.

Therefore, the correct answer given the choices is:

A) F(x) is a negative number with a large absolute value

We know this because the absolute value of -infinity is infinity!

I hope this explanation made sense.

answer: the answer is 23

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}