Find the values of a and b that make f continuous everywhere.
f(x) =
(x2 − 4)/(x − 2) if x < 2
ax2 − bx + 3 if 2 ≤ x < 3
4x − a + b if x ≥ 3
solve for a and b? HELP ive tried everything
To find the values of a and b that make f continuous everywhere, we need to ensure that the different expressions for f(x) match up at the transition points.
First, let's consider the left-hand limit of f(x) as x approaches 2. We can find this limit by evaluating the expression f(x) = (x^2 - 4)/(x - 2) as x approaches 2 from the left side:
lim(x→2-) [(x^2 - 4)/(x - 2)]
To simplify, we can factor the numerator:
lim(x→2-) [(x + 2)(x - 2)/(x - 2)]
Canceling out the common factors:
lim(x→2-) [(x + 2)]
Taking the limit as x approaches 2, we find that the left-hand limit is 4.
Now, let's consider the right-hand limit of f(x) as x approaches 2. We can evaluate the expression f(x) = ax^2 - bx + 3 as x approaches 2 from the right side:
lim(x→2+) [ax^2 - bx + 3]
Plugging in the value x = 2:
a(2^2) - b(2) + 3 = 4a - 2b + 3
The right-hand limit should also be equal to 4.
Now, let's consider the left-hand limit of f(x) as x approaches 3. We can evaluate the expression f(x) = ax^2 - bx + 3 as x approaches 3 from the left side:
lim(x→3-) [ax^2 - bx + 3]
Similarly, for the right-hand limit of f(x) as x approaches 3, we need to evaluate the expression f(x) = 4x - a + b as x approaches 3 from the right side:
lim(x→3+) [4x - a + b]
Both these limits should be equal to each other.
To summarize, we have the following equations:
lim(x→2-) [(x + 2)] = 4
lim(x→2+) [(4a - 2b + 3)] = 4
lim(x→3-) [(ax^2 - bx + 3)] = lim(x→3+) [(4x - a + b)]
Now, using these equations, we can solve for the values of a and b by substituting the corresponding limits:
lim(x→2-) [(2 + 2)] = 4
4 = 4
lim(x→2+) [(4a - 2b + 3)] = 4
4a - 2b + 3 = 4
lim(x→3-) [(ax^2 - bx + 3)] = lim(x→3+) [(4x - a + b)]
a(3^2) - b(3) + 3 = 4(3) - a + b
Now, you can solve these equations to find the values of a and b.