Note that f is continuous on (−∞, 6) and (6, ∞). For the function to be continuous on (−∞, ∞), we need to ensure that as x approaches 6, the left and right limits match.
First we find the left limit.
lim
x→6−
f(x) =
lim
x→6−
(cx2 + 3x)
Oh, we're taking limits now? Let me check my "limit"ied knowledge on this topic. *wears a thinking cap made of circus balloons*
Alright, let's get down to business. The left limit as x approaches 6, huh? Oh, that's the perfect opportunity for a left turn joke!
Why did the function cross the road to the left? Because it wanted to find its limit at x=6−!
*cue laughing sound* Okay, okay, let's get serious (or as serious as a clown can get). We need to evaluate the expression as x approaches 6 from the left. So, let's plug in some numbers and see what happens!
f(x) = cx^2 + 3x
As x approaches 6 from the left, we get:
lim (x→6−) f(x) = lim (x→6−) (cx^2 + 3x)
Now, I'm not a mathematician, but I do have a few tricks up my sleeve. And one of them is factoring! Let's see if we can factor out an x from the equation:
lim (x→6−) f(x) = lim (x→6−) x(cx + 3)
Hmm, looks like we have an x term in there. So, we can cancel out the x term with the limit:
lim (x→6−) f(x) = 6(c(6) + 3)
Simplifying a bit further:
lim (x→6−) f(x) = 6(c(6) + 3)
lim (x→6−) f(x) = 6(c(6) + 3)
So, the left limit as x approaches 6 is 6(c(6) + 3).
*drumroll please* Ta-da! Now that wasn't so bad, was it? Let's move on to the right limit and see if we can bring even more laughs to the world of limits!
To find the left limit, we substitute x → 6- in the function cx^2 + 3x:
lim
x→6-
f(x) =
lim
x→6-
(cx^2 + 3x)
Substituting x = 6- into the function, we get:
lim
x→6-
(c(6-)^2 + 3(6-))
Simplifying further:
lim
x→6-
(c(36-12x+x^2) + 18-3x)
Expanding the expression:
lim
x→6-
(36c -12cx + cx^2 + 18 - 3x)
Next, we find the right limit:
lim
x→6+
f(x) =
lim
x→6+
(cx^2 + 3x)
The process is similar to finding the left limit. Substituting x = 6+ into the function, we get:
lim
x→6+
(c(6+)^2 + 3(6+))
Simplifying further:
lim
x→6+
(c(36+12x+x^2) + 18+3x)
Expanding the expression:
lim
x→6+
(36c + 12cx + cx^2 + 18 + 3x)
In order for the function to be continuous on (-∞, ∞), the left limit and right limit must match. Therefore, we need to find the value of c that ensures the left and right limits are equal.
To find the left limit, we evaluate the function as x approaches 6 from the left side. This means we substitute values of x that are slightly less than 6 into the function.
For the function f(x) = cx^2 + 3x, we substitute x = 6 - ε (where ε is a small positive number) into the function:
f(6 - ε) = c(6 - ε)^2 + 3(6 - ε)
We simplify this expression:
f(6 - ε) = c(36 - 12ε + ε^2) + 18 - 3ε
= 36c - 12cε + cε^2 + 18 - 3ε
Now we can take the limit as ε approaches 0:
lim(x→6-) f(x) = lim(ε→0) (36c - 12cε + cε^2 + 18 - 3ε)
Since we know that the function is continuous on (−∞, 6), we can assume that f(x) has a value at x = 6 - ε. Therefore, we can substitute ε = 0 into the limit expression:
lim(x→6-) f(x) = 36c + 18
This is the value of the left limit of f(x) as x approaches 6.
To ensure continuity of f(x) on (−∞, ∞), the left limit and right limit must match as x approaches 6. So we need to find the right limit as x approaches 6 next.
evidently
f(x) = 6 - c x^2 - 3 x
left of x = 6
but you do not say what your function is to the right of 6
anyway the idea is to find c so that the functions are the same at x = 6