For what values of a and b is
F(x) = {ax+2b , xless than 0
{X²+3a-b , 0 less than x less than 2
{3x-5 , x greather than continous at every x
To ensure continuity at every x, the value of the function at x = 0 and x = 2 must be equal. So, we have:
ax + 2b = 0² + 3a - b
ax + 2b = 3a - b
Since these expressions must be equal, the coefficients of x and constant terms on both sides must be equal.
Equating the coefficients of x:
a = 3
Equating the constants:
2b = -b
3b = 0
b = 0
Therefore, for the function to be continuous at every x, a = 3 and b = 0.
To ensure that the function F(x) is continuous at every x, we need to make sure that the pieces of the function "join smoothly" at the points of transition, namely x = 0 and x = 2.
At x = 0, we have:
F(0) = 0² + 3a - b (using the second piece of the function)
To have continuity at x = 0, this value must be equal to the value of the first piece of the function at x = 0: F(0) = 2b.
Therefore, we have the equation:
0² + 3a - b = 2b
This simplifies to:
3a - 2b + b = 0
3a - b = 0
This is the first condition for continuity.
At x = 2, we have:
F(2) = 3(2) - 5 (using the third piece of the function)
To have continuity at x = 2, this value must be equal to the value of the second piece of the function at x = 2: F(2) = 2² + 3a - b.
Therefore, we have the equation:
3(2) - 5 = 2² + 3a - b
6 - 5 = 4 + 3a - b
1 = 4 + 3a - b
This simplifies to:
3a - b = -3
This is the second condition for continuity.
To summarize, for F(x) to be continuous at every x, we need to solve the following system of equations:
3a - b = 0 (Equation 1)
3a - b = -3 (Equation 2)
Subtracting Equation 2 from Equation 1, we get:
0 = 3
This is a contradiction. There are no values of a and b that satisfy both equations simultaneously, and therefore, there are no values of a and b for which the function F(x) is continuous at every x.