The function f is given by the formula: f(x)= (4x^3 + 24x^2 + 9x + 54)/(x+6) when x < -6 and by the formula: 4x^2 + 1x + A when -6 <= (greater than or equal) x. What value must be chosen...

The function f is given by the formula: 

f(x)= (4x^3 + 24x^2 + 9x + 54)/(x+6) when x < -6 and by the formula: 
4x^2 + 1x + A 

when -6 <= (greater than or equal) x. 

What value must be chosen for a in order to make this function continuous at -6?

Asked on by GFroun

1 Answer | Add Yours

tiburtius's profile pic

tiburtius | High School Teacher | (Level 2) Educator

Posted on

First you need to find limit of `f(x)` when `x->-6.`

`lim_(x->-6)(4x^3+24x^2+9x+54)/(x+6)=`

Now we divide `4x^3+24x^2+9x+54` by `x-6` so we get

`lim_(x->-6)(4x^2+9)=4(-6)^2+9=153`  

So the limit of this function at -6 is 153 so in order to make the function continuous at -6 the other part of the function must also have value of 153 at x=-6. That is when we put x=-6 in `4x^2+x+A` we must get 153.

`4(-6)^2-6+A=153`

`144-6+A=153`

` `

`A=15` <-- A must be equal to 15.

Your function is 

`f(x)={((4x^3+24x^2+9x+54)/(x+6) if x<-6),(4x^2+x+15 if x>=-6):}`

Graph:

Blue and red curve meet at x=-6.

We’ve answered 318,928 questions. We can answer yours, too.

Ask a question