Otwo decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x= 3(x) + k for x less than or equal to 3 and is equal to k(x) squared - 6 for x greater than 3

  k ≈ 1.88

Step-by-step explanation:

In order for the function to be continuous, the pieces of the function must have the same value at x=3. That is ...

  f(3) = 3(3) +k = k(3^2) -6

  15 = 8k . . . . . add 6-k, simplify

  k = 15/8 = 1.875 ≈ 1.88

To two decimal places, the value of k that makes f(x) continuous at x=3 is 1.88.

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}

y > 1/3x

y < = -x + 4

hope this : )