Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)
x^2-4/x-2 x<2
f(x)= ax^2-bx+5 2¡Üx<3
2x-a+b x¡Ý3
a=
b=
To find the values of a and b that make the function f continuous everywhere, we need to make sure that the function is continuous at the points where the different parts of the function meet.
First, let's consider the point x = 2, which is the end point of the first part of the function and the starting point of the second part.
For the function to be continuous at x = 2, the values of the first part and the second part of the function should match at this point.
So, we need to find the limit of f(x) as x approaches 2 from the left (x → 2⁻) and as x approaches 2 from the right (x → 2⁺), and then set them equal to each other.
Limit as x → 2⁻ of f(x):
lim(x → 2⁻) (x^2 - 4)/(x - 2) = lim(x → 2⁻) [(x - 2)(x + 2)/(x - 2)] = lim(x → 2⁻) (x + 2) = 2 + 2 = 4
Limit as x → 2⁺ of f(x):
lim(x → 2⁺) (ax^2 - bx + 5) = a(2)^2 - b(2) + 5 = 4a - 2b + 5
Setting the limits equal to each other:
4 = 4a - 2b + 5
Simplifying the equation:
-1 = 4a - 2b
Now, let's consider the point x = 3, which is the end point of the second part and the starting point of the third part.
For the function to be continuous at x = 3, the values of the second part and the third part of the function should match at this point.
So, we need to find the limit of f(x) as x approaches 3 from the left (x → 3⁻) and as x approaches 3 from the right (x → 3⁺), and then set them equal to each other.
Limit as x → 3⁻ of f(x):
lim(x → 3⁻) (ax^2 - bx + 5) = a(3)^2 - b(3) + 5 = 9a - 3b + 5
Limit as x → 3⁺ of f(x):
lim(x → 3⁺) (2x - a + b) = 2(3) - a + b = 6 - a + b
Setting the limits equal to each other:
9a - 3b + 5 = 6 - a + b
Simplifying the equation:
10a - 4b + 5 = 6
Rearranging the terms:
10a - 4b = 1
Now, we have a system of two equations with two unknowns:
-1 = 4a - 2b (Equation 1)
10a - 4b = 1 (Equation 2)
Solving this system of equations, we can find the values of a and b.
Multiplying Equation 1 by 5, we get:
-5 = 20a - 10b (Equation 3)
Adding Equation 2 and Equation 3, we eliminate the b term:
10a - 4b + (-5) = 1 + (-5)
10a - 4b - 5 = -4
Rearranging the terms:
10a - 4b = -4 + 5
10a - 4b = 1
Comparing this equation to Equation 2, we can see that they are the same:
10a - 4b = 1
10a - 4b = 1
This means that the system of equations is dependent, and there are infinitely many possible solutions.
To find a particular solution, we can set one of the variables to a known value. Let's set b = 0.
Plugging b = 0 into Equation 1, we get:
-1 = 4a - 2(0)
-1 = 4a
a = -1/4
Now that we have the value of a, we can find the value of b from Equation 1:
-1 = 4(-1/4) - 2b
-1 = -1 - 2b
-1 + 1 = -2b
0 = -2b
b = 0
Therefore, the values of a and b that make f continuous everywhere are:
a = -1/4
b = 0
To find the values of a and b that make the function f continuous everywhere, we need to ensure that f(x) is continuous at x = 2 and x = 3.
First, let's find the limit of f(x) as x approaches 2 from the left (x < 2):
lim(x→2-) f(x) = lim(x→2-) (x^2-4)/(x-2)
To evaluate this limit, we can simplify the expression:
lim(x→2-) (x^2-4)/(x-2) = lim(x→2-) [(x-2)(x+2)/(x-2)]
Since (x-2) cancels out, we are left with:
lim(x→2-) (x+2) = 2 + 2 = 4
Now, let's find the limit of f(x) as x approaches 2 from the right (2 ≤ x < 3):
lim(x→2+) f(x) = lim(x→2+) (ax^2-bx+5)
To find the limit, we substitute x = 2 into the expression:
lim(x→2+) f(x) = a(2^2)-b(2)+5 = 4a - 2b + 5
To make sure the function is continuous at x = 2, the limit from the left (4) must be equal to the limit from the right (4a - 2b + 5):
4 = 4a - 2b + 5
Simplifying the equation, we get:
4a - 2b = -1 (Equation 1)
Next, let's find the limit of f(x) as x approaches 3 from the left (2 ≤ x < 3):
lim(x→3-) f(x) = lim(x→3-) (ax^2-bx+5)
Substituting x = 3 into the expression:
lim(x→3-) f(x) = a(3^2)-b(3)+5 = 9a - 3b + 5
Now, let's find the limit of f(x) as x approaches 3 from the right (x ≥ 3):
lim(x→3+) f(x) = lim(x→3+) (2x-a+b)
Substituting x = 3 into the expression:
lim(x→3+) f(x) = 2(3)-a+b = 6-a+b
To ensure the function is continuous at x = 3, the limit from the left (9a - 3b + 5) must be equal to the limit from the right (6 - a + b):
9a - 3b + 5 = 6 - a + b
Simplifying the equation, we get:
10a - 4b = 1 (Equation 2)
Now, we have two equations (Equation 1 and Equation 2) with two variables (a and b). We can solve this system of equations to find the values of a and b.
Multiplying Equation 1 by 2, we get:
8a - 4b = -2 (Equation 3)
Adding Equation 3 to Equation 2, we can eliminate b:
(8a - 4b) + (10a - 4b) = -2 + 1
18a = -1
a = -1/18
Substituting the value of a back into Equation 1, we can solve for b:
4(-1/18) - 2b = -1
-4/18 - 2b = -1
-2/9 - 2b = -1
-2b = -1 + 2/9
-2b = -9/9 + 2/9
-2b = -7/9
b = -(-7/9) / 2
b = 7/18
Therefore, the values of a and b that make f continuous everywhere are:
a = -1/18
b = 7/18