Determine the values of a and b to make the following function continuous at every value of x.?

f(x)= { e^(x^2-x+a) if x < 1 , (4^x-x^2)) if 1<x<2 , 6In(x-b) if x>2

e^(x^2-x+a) if x < 1

as x->1-, f(x) -> e^a

(4^x-x^2)) if 1<x<2
as x->1+, f(x) -> 4^1-1 = 3
as x->2-, f(x) -> 4^2-2^2 = 12

6In(x-b) if x>2
as x->2-, f(x) -> 6ln(2-b)

so, we need

e^a = 3
a = ln3

6ln(2-b) = 12
ln(2-b)=2
2-b = e^2
b = 2-e^2

But using those values will not make f(x) continuous at every x, since f(x) is not defined at x=1 or x=2. To fix that, we need to add

f(1) = 3
f(2) = 12

Or make some of the intervals closed on one end or or maybe both ends.

To make the function continuous at every value of x, we need to ensure that the function is defined and has the same value at the point where the different pieces of the function meet. In this case, we have three different pieces of the function for x < 1, 1 < x < 2, and x > 2.

To determine the values of a and b, we'll examine the conditions at the points where the different pieces of the function meet. Let's analyze each piece separately:

1. For x < 1:
The function is given by f(x) = e^(x^2 - x + a).
We don't need to worry about continuity at x = 1 for this piece since there is no point of intersection. Therefore, we don't have any conditions for a to make the function continuous at x < 1.

2. For 1 < x < 2:
The function is given by f(x) = (4^x - x^2).
To make the function continuous at x = 1, we need to ensure that the two pieces (x < 1 and 1 < x < 2) match at x = 1.
Taking the left-hand limit as x approaches 1, we have:
lim(x→1-) (4^x - x^2) = 4^1 - 1^2 = 4 - 1 = 3.
Therefore, the value of the function at x = 1 is f(1) = 3.
However, the function given for x < 1 is f(x) = e^(x^2 - x + a), so we need to find the value of this function at x = 1 to check if it matches f(1) = 3.
Plugging x = 1 into f(x) = e^(x^2 - x + a), we get:
f(1) = e^(1^2 - 1 + a) = e^(1 - 1 + a) = e^a.
To make the function continuous at x = 1, e^a must be equal to 3. Therefore, we have:
e^a = 3 (Equation 1).

3. For x > 2:
The function is given by f(x) = 6 ln(x - b).
To make the function continuous at x = 2, we need to ensure that the two pieces (1 < x < 2 and x > 2) match at x = 2.
Taking the right-hand limit as x approaches 2, we have:
lim(x→2+) 6ln(x - b) = 6ln(2 - b).
Therefore, the value of the function at x = 2 is f(2) = 6 ln(2 - b).
However, the function given for 1 < x < 2 is f(x) = (4^x - x^2), so we need to find the value of this function at x = 2 to check if it matches f(2) = 6 ln(2 - b).
Plugging x = 2 into f(x) = (4^x - x^2), we get:
f(2) = (4^2 - 2^2) = 16 - 4 = 12.
To make the function continuous at x = 2, 6 ln(2 - b) must be equal to 12. Therefore, we have:
6 ln(2 - b) = 12 (Equation 2).

Now, we have two equations (Equation 1 and Equation 2) and two unknowns (a and b). We can solve this system of equations to find the values of a and b that make the function continuous at every value of x.