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 the function f(x) continuous, we need to ensure that the function has the same value from both sides when x = 2. In other words, we need to make sure that the two parts of the function, defined for x <= 2 and x < 2, agree at x = 2.
We'll start by evaluating f(x) from both sides when x = 2.
For x <= 2:
f(x) = x^2 + 3x - 1
f(2) = 2^2 + 3(2) - 1
= 4 + 6 - 1
= 9
For x < 2:
f(x) = -3bx + 3
f(2) = -3b(2) + 3
= -6b + 3
To make the function continuous at x = 2, the values from both sides must match. Hence, we have:
9 = -6b + 3
Now, let's solve this equation to determine the value of b:
9 - 3 = -6b
6 = -6b
b = -1
Therefore, the value of b that makes the function f(x) continuous is b = -1.