If f(x) = {x^2+3x-1, x<=2 and -3bx+3, x<2 , find the value of b in order for f to be continuous
To find the value of b that makes f(x) continuous, we need to ensure that the two pieces of the function f(x) intersect at x = 2. In other words, we need to make sure that the limit of f(x) as x approaches 2 is equal to the value of f(x) when x is exactly 2.
To determine the limit as x approaches 2, we can take the limit of each piece of f(x) separately.
For the first piece of f(x), which is x^2 + 3x - 1, we can plug in x = 2:
lim(x→2) (x^2 + 3x - 1) = (2^2 + 3*2 - 1) = 9
So, the limit of f(x) as x approaches 2 from the left side is 9.
For the second piece of f(x), which is -3bx + 3, we need to evaluate it as x approaches 2 from the right side. Since x < 2 for this piece, we should plug in x = 2 - δ, where δ is a small positive number approaching zero:
lim(x→2+) (-3bx + 3) = (-3b(2 - δ) + 3) = -6b + 3
Now, to make the function f(x) continuous at x = 2, we need to equate the two limits:
9 = -6b + 3
Solving this equation for b, we can add 6b to both sides and subtract 3 from both sides:
6b = 9 - 3
6b = 6
b = 1
Therefore, the value of b that makes f(x) continuous is b = 1.