For what value of the constant c is the function f continuous on the interval (-infinity, +infinity).
f(x) = cx^2+4x, x < 2
f(x) = x^3-cx
limit on the left is 4c+8
limit on the right is 8-2c
so you need
4c+8 = 8-2c
c=0
maybe not the best exercise for the topic, yeah?
To find the value of the constant c for which the function f is continuous on the interval (-∞, +∞), you need to check if the two pieces of the function, f(x) = cx^2 + 4x, for x < 2 and f(x) = x^3 - cx, match each other at the point x = 2.
First, let's determine the left-hand limit as x approaches 2. To do this, substitute x = 2 into the first piece of the function:
lim(x→2-) (cx^2 + 4x) = c(2^2) + 4(2) = 4c + 8
Next, let's determine the right-hand limit as x approaches 2. To do this, substitute x = 2 into the second piece of the function:
lim(x→2+) (x^3 - cx) = 2^3 - c(2) = 8 - 2c
For the function to be continuous at x = 2, the left-hand limit and the right-hand limit must be equal:
4c + 8 = 8 - 2c
Now, solve this equation for c:
4c + 2c = 8 - 8
6c = 0
c = 0
Therefore, the constant c = 0 makes the function f continuous on the interval (-∞, +∞).