The function f(x)=x+3x2−9 is defined everywhere except at x=±3. if possible, define f at x=−3 so that it becomes continuous at x=−3.

The function simplifies to

By defining f(-3) to be 1/(-3-3) = -1/6, we can plug the "hole" at that point, making the function continuous there.

_____

The attachments show the original function being undefined at x=-3. Defining the function at x=-3 removes the discontinuity.

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5