Help with calculus Continuity problem?

Question

f(x)={(x^2+6x+8)/(x+2) if x<-2
and {kx^2 if x> or eqaul to -2

What value of k would make the function continuous at x=-2

Answer 1

f(x) = { (x^2 + 6x + 8) / (x + 2) if x < -2
{ kx^2 if x >= -2

In order for this function to be continuous, these three values must be equal to each other:

f(-2)

lim f(x) {x approaches -2 from the right}
x -> -2+

lim f(x) {x approaches -2 from the left}
x -> -2-

First, let's solve f(-2). By the piecewise function
f(-2) = k(-2)^2 {since x >= -2}, so
f(-2) = 4k

Now, for

lim f(x)
x -> -2-

We choose the first function, since the limit is being taken from the left (that is, x < -2) and close to _but not equal to_ -2. Thus, we have

lim [(x^2 + 6x + 8) / (x + 2)]
x -> (-2)-

Factor the top,

lim [(x + 2) (x + 4) / (x + 2)]
x -> (-2)-

Cancel (x + 2) on the top and bottom,

lim [x + 4]
x -> (-2)-

and now we can plug x = -2 directly, giving us
-2 + 4 = 2

And, by the definition of continuity, this must equal f(-2) = 4k. So

4k = 2, meaning
k = 2/4 = 1/2

The value of k that would make the function continuous at x = -2 is 1/2.

Answer 2

you have to take the lim when x->-2 of (x^2+6x+8)/(x+2)
which is 0/0, so you can take l'Hopital's rule and get f'(x)= (2x+6)/1 and the lim of that when x-> = -4+6 = 2
So now you need that lim when x->-2 of kx^2=2
=> k(-2)^2=2
=> k=2/4
=> k=1/2 which is the answer

Answer 3

We see that on (-oo,pi/4), sin(x) is non-end, to boot as cos(x) on (pi/4,oo). So, the only factor at which continuity is a controversy is pi/4. We in many situations do those by skill of exhibiting the left and spectacular limits of f(x) are equivalent and equivalent to f(pi/4), which skill does sin(x) = cos(x) at pi/4. If it does, then the function's curve remains linked and (you guessed it, non-end). properly cos(pi/4) = one million/sqrt(2) and sin(pi/4)=one million/sqrt(2). Yay! that's non-end. in case you ought to use limits, then instruct that lim sin(x) = 1sqrt(2) as x procedures pi/4 from the left and lim cos(x) = 1sqrt(2) as x approachs pi/4 from the spectacular.

Answer 4

0 i think