Find the constants a and b such that the function

​

f(x)= 4, x&lt;=negative 1; f(x)= ax+b, negative 1&lt;x&lt;3; f(x)=negative 4, x&gt;=3

​

is continuous on the entire real line.

To ensure the function \( f(x) \) is continuous on the entire real line, the function must not have any jumps or discontinuities at the points where the piecewise function changes its definition, specifically at \( x = -1 \) and \( x = 3 \).

Given:
\[
f(x) = \begin{cases}
4, & x \leq -1 \\
ax + b, & -1 < x < 3 \\
-4, & x \geq 3
\end{cases}
\]

### Step 1: Continuity at \( x = -1 \)
For the function to be continuous at \( x = -1 \), the left-hand limit (as \( x \) approaches \( -1 \) from the left) must equal the right-hand limit (as \( x \) approaches \( -1 \) from the right), and both must equal \( f(-1) \).

- Left-hand limit as \( x \) approaches \( -1 \) from the left:
  \[
  \lim_{x \to -1^-} f(x) = 4
  \]
- Right-hand limit as \( x \) approaches \( -1 \) from the right:
  \[
  \lim_{x \to -1^+} f(x) = a(-1) + b = -a + b
  \]
- For continuity at \( x = -1 \):
  \[
  \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1)
  \]
  \[
  4 = -a + b
  \]

### Step 2: Continuity at \( x = 3 \)
For the function to be continuous at \( x = 3 \), the left-hand limit (as \( x \) approaches \( 3 \) from the left) must equal the right-hand limit (as \( x \) approaches \( 3 \) from the right), and both must equal \( f(3) \).

- Left-hand limit as \( x \) approaches \( 3 \) from the left:
  \[
  \lim_{x \to 3^-} f(x) = a(3) + b = 3a + b
  \]
- Right-hand limit as \( x \) approaches \( 3 \) from the right:
  \[
  \lim_{x \to 3^+} f(x) = -4
  \]
- For continuity at \( x = 3 \):
  \[
  \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)
  \]
  \[
  3a + b = -4
  \]

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

1. \(-a + b = 4\)
2. \(3a + b = -4\)

Subtract equation 1 from equation 2 to eliminate \( b \):

\[
(3a + b) - (-a + b) = -4 - 4
\]
\[
4a = -8
\]
\[
a = -2
\]

Now, substitute \( a = -2 \) into equation 1 to solve for \( b \):

\[
-a + b = 4
\]
\[
-(-2) + b = 4
\]
\[
2 + b = 4
\]
\[
b = 2
\]

### Final Answer
The constants are \( a = -2 \) and \( b = 2 \).

Would you like further details or have any questions? 

Here are 5 related questions you might find interesting:
1. How can you determine if a function is continuous at a specific point?
2. What is the difference between continuity and differentiability?
3. How would the answer change if the function \( f(x) \) was defined differently for \( x \geq 3 \)?
4. What are the conditions for a piecewise function to be continuous everywhere?
5. How would you approach finding the limits of more complex piecewise functions?

**Tip:** When solving for continuity in piecewise functions, always equate the limits at the points where the function definition changes to ensure smooth transitions.

Find the constants a and b such that the function f(x)= 4, x<=-1; f(x)= ax+b, -1<x<3; f(x)=-4, x>=3 is continuous on the entire real line.

Determine the constants a and b for a piecewise function to ensure it is continuous across the entire real line. This problem covers algebraic concepts, including continuity and piecewise functions, suitable for Grades 11-12 students.

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