what values of a and b make f(x) continous everywhere for f(x)={x^2-4/x-2 for x<2, ax^2-bx-30 for 2<=x<3 16x-a+b for 3<=x

To find the values of \(a\) and \(b\) that make the function \(f(x)\) continuous everywhere, we need to ensure continuity at the points \(x = 2\) and \(x = 3\). Specifically, the function must have the same value from both the left and right at these points.

The function \(f(x)\) is defined piecewise as follows:

\[
f(x) = \begin{cases} 
\frac{x^2 - 4}{x - 2} & \text{for } x < 2, \\
ax^2 - bx - 30 & \text{for } 2 \leq x < 3, \\
16x - a + b & \text{for } 3 \leq x.
\end{cases}
\]

### Step 1: Continuity at \(x = 2\)
For the function to be continuous at \(x = 2\), the limit from the left (\(x \to 2^{-}\)) must equal the limit from the right (\(x \to 2^{+}\)), and the function must be defined at \(x = 2\).

#### Left-hand limit as \(x \to 2^-\):
For \(x < 2\), we have \(f(x) = \frac{x^2 - 4}{x - 2}\). Simplifying this expression:
\[
\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for } x \neq 2.
\]
So, as \(x \to 2^-\), the limit of \(f(x)\) is:
\[
\lim_{x \to 2^-} f(x) = 2 + 2 = 4.
\]

#### Right-hand limit as \(x \to 2^+\):
For \(x \geq 2\), we have \(f(x) = ax^2 - bx - 30\). To ensure continuity at \(x = 2\), we require:
\[
a(2)^2 - b(2) - 30 = 4.
\]
Simplifying:
\[
4a - 2b - 30 = 4.
\]
Thus, the first equation is:
\[
4a - 2b = 34. \quad \text{(Equation 1)}
\]

### Step 2: Continuity at \(x = 3\)
For the function to be continuous at \(x = 3\), the limit from the left (\(x \to 3^{-}\)) must equal the limit from the right (\(x \to 3^{+}\)).

#### Left-hand limit as \(x \to 3^-\):
For \(2 \leq x < 3\), we have \(f(x) = ax^2 - bx - 30\). So, as \(x \to 3^{-}\), the limit of \(f(x)\) is:
\[
a(3)^2 - b(3) - 30 = 9a - 3b - 30.
\]

#### Right-hand limit as \(x \to 3^+\):
For \(x \geq 3\), we have \(f(x) = 16x - a + b\). So, as \(x \to 3^{+}\), the limit of \(f(x)\) is:
\[
16(3) - a + b = 48 - a + b.
\]
To ensure continuity at \(x = 3\), we require:
\[
9a - 3b - 30 = 48 - a + b.
\]
Simplifying:
\[
9a - 3b + a - b = 78.
\]
Thus, the second equation is:
\[
10a - 4b = 78. \quad \text{(Equation 2)}
\]

### Step 3: Solving the system of equations
We now have the following system of equations:

1. \(4a - 2b = 34\),
2. \(10a - 4b = 78\).

To solve this system, we can use substitution or elimination. I'll solve it using the elimination method.

#### Eliminate \(b\):
Multiply the first equation by 2:
\[
8a - 4b = 68.
\]
Now subtract this equation from the second equation:
\[
(10a - 4b) - (8a - 4b) = 78 - 68,
\]
which simplifies to:
\[
2a = 10.
\]
Thus, \(a = 5\).

#### Solve for \(b\):
Substitute \(a = 5\) into the first equation:
\[
4(5) - 2b = 34,
\]
which simplifies to:
\[
20 - 2b = 34,
\]
so \(2b = -14\), and therefore \(b = -7\).

### Final Answer:
The values of \(a\) and \(b\) that make \(f(x)\) continuous everywhere are:
\[
a = 5, \quad b = -7.
\]

Would you like more details on the solution, or do you have any questions?

Here are 5 related questions:
1. How do you check for continuity at other points in a piecewise function?
2. What is the general procedure for solving systems of equations?
3. How does differentiability relate to continuity in piecewise functions?
4. What happens if a function is not continuous at a point?
5. Can a function be continuous but not differentiable?

**Tip**: When solving piecewise functions, always check for both limits and equality of function values at the boundary points.

what values of a and b make f(x) continuous everywhere for f(x) = { (x^2 - 4) / (x - 2) for x < 2, ax^2 - bx - 30 for 2 <= x < 3, 16x - a + b for 3 <= x }

Learn how to find the values of a and b that make a piecewise function continuous everywhere. This problem involves limits, continuity, and solving systems of equations, ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation