The graph of a secant function shows no functional values on (−7, 7) . The asymptotes of the function occur every π units. Which function represents the secant function described?

f(x)=sec(x/2)+7

f(x)=7sec(x/2)

f(x)=7sec(x)

f(x)=sec(x)+7

f(x) = 7 sec(x)

Step-by-step explanation:

The range of the function is less than -7 or greater than 7.  So the midline is y=0 and the amplitude is 7.

There are two asymptotes per period, so the period of the function is 2π.

Therefore, the function is:

f(x) = A sec(2π / T x)

f(x) = 7 sec(2π / (2π) x)

f(x) = 7 sec(x)

  f(x) = 7sec(x)

Step-by-step explanation:

The parent secant function has an excluded range of (-1, 1), so the one we're seeking has been scaled vertically by a factor of 7. It will look like ...

  f(x) = 7sec( )

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The parent secant function has vertical asymptotes at odd multiples of π/2, so are separated by π units. Thus, there is no horizontal scaling.

The function described is f(x) = 7sec(x).

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}

20

step-by-step explanation: