Let f(x)=x2−12x. To prove that limx→6f(x)=−36, we proceed as follows. Given any ϵ>0, we need to find a number δ>0 such that if 0<|x−6|<δ, then |(x2−12x)−(−36)|<ϵ. What is the (largest) choice of δ that is certain to work? (Your answer will involve ϵ. When entering your answer, type e in place of ϵ.)

Take

Step-by-step explanation:

To begin notice that

If we chose the following delta

that solves our problem !

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.