For what value of the constant c is the function f continuous on (−∞, ∞)?
f(x) =
cx2 + 4x if x < 3
x3 − cx if x ≥ 3
To find the value of the constant c for which the function f is continuous on (-∞, ∞), we need to ensure that the two pieces of the function, f(x) = cx^2 + 4x for x < 3 and f(x) = x^3 - cx for x ≥ 3, meet at the point x = 3 without any discontinuity.
To do this, we need to find the limit of f(x) as x approaches 3 from both the left and the right side and make sure they are equal.
Limit as x approaches 3 from the left side:
lim(x→3-) cx^2 + 4x
Evaluating this limit gives us:
9c + 12
Limit as x approaches 3 from the right side:
lim(x→3+) x^3 - cx
Evaluating this limit gives us:
27 - 3c
For the function to be continuous at x = 3, the limits from the left and right sides must be equal, so we have the equation:
9c + 12 = 27 - 3c
Simplifying this equation gives us:
12c + 12 = 27
12c = 27 - 12
12c = 15
c = 15/12
c = 5/4
Therefore, the value of the constant c that makes the function f continuous on (-∞, ∞) is c = 5/4.
To determine the value of the constant c for which the function f is continuous on (−∞, ∞), we need to ensure that the values of f(x) from the left and right of x = 3 match up.
First, let's find the limit of f(x) as x approaches 3 from the left side (x < 3):
lim(x→3-) f(x) = lim(x→3-) (cx^2 + 4x)
Next, let's find the limit of f(x) as x approaches 3 from the right side (x ≥ 3):
lim(x→3+) f(x) = lim(x→3+) (x^3 - cx)
For the function to be continuous at x = 3, these two limits should be equal.
Setting up the limits:
lim(x→3-) (cx^2 + 4x) = lim(x→3+) (x^3 - cx)
Now, let's calculate the limits:
lim(x→3-) (cx^2 + 4x) = c(3^2) + 4(3) = 9c + 12
lim(x→3+) (x^3 - cx) = 3^3 - c(3) = 27 - 3c
For continuity, these two limits should be equal. Therefore:
9c + 12 = 27 - 3c
Simplifying the equation:
12c + 12 = 27
Subtracting 12 from both sides:
12c = 15
Dividing both sides by 12:
c = 15/12 = 5/4
So, the function f(x) will be continuous on (−∞, ∞) if the constant c is equal to 5/4.
There's some font mangling from your copy/paste, but if I read it right,
f(x) =
cx^2 if x < 3
x^3 if x >= 3
limit(x->3-) = 9c
limit(x->3+) = 27
so, if c=3 the two branches join at (3,27)