Math Problem Statement

Find the values of a and b that make f continuous everywhere.

Solution

To solve for the values of aa and bb that make the function f(x)f(x) continuous everywhere, we need to ensure continuity at the points where the function pieces meet, specifically at x=8x = -8 and x=1x = -1.

Step 1: Continuity at x=8x = -8

For f(x)f(x) to be continuous at x=8x = -8, the left-hand limit (from the first piece) must equal the right-hand limit (from the second piece), and both must equal f(8)f(-8).

  1. Left-hand limit at x=8x = -8: The function for x<8x < -8 is: f(x)=x264x+8f(x) = \frac{x^2 - 64}{x + 8} Simplifying this expression: f(x)=(x8)(x+8)x+8=x8forx8f(x) = \frac{(x-8)(x+8)}{x+8} = x - 8 \quad \text{for} \quad x \neq -8 So, as x8x \to -8^-: limx8f(x)=88=16\lim_{x \to -8^-} f(x) = -8 - 8 = -16

  2. Right-hand limit at x=8x = -8: The function for 8x<1-8 \leq x < -1 is: f(x)=ax2bx448f(x) = ax^2 - bx - 448 Substituting x=8x = -8: f(8)=a(8)2b(8)448=64a+8b448f(-8) = a(-8)^2 - b(-8) - 448 = 64a + 8b - 448 To ensure continuity, we set the left-hand limit equal to the right-hand value: 64a+8b448=1664a + 8b - 448 = -16 Simplifying: 64a+8b=432(Equation 1)64a + 8b = 432 \quad \text{(Equation 1)}

Step 2: Continuity at x=1x = -1

For f(x)f(x) to be continuous at x=1x = -1, the second piece (for 8x<1-8 \leq x < -1) and the third piece (for x1x \geq -1) must match at x=1x = -1.

  1. Left-hand limit at x=1x = -1: Using the second piece ax2bx448ax^2 - bx - 448, substitute x=1x = -1: f(1)=a(1)2b(1)448=a+b448f(-1) = a(-1)^2 - b(-1) - 448 = a + b - 448

  2. Right-hand limit at x=1x = -1: The function for x1x \geq -1 is 436xa+b436x - a + b. Substituting x=1x = -1: f(1)=436(1)a+b=436a+bf(-1) = 436(-1) - a + b = -436 - a + b Setting the left-hand limit equal to the right-hand limit for continuity: a+b448=436a+ba + b - 448 = -436 - a + b Simplifying: 2a=12a=6(Equation 2)2a = 12 \quad \Rightarrow \quad a = 6 \quad \text{(Equation 2)}

Step 3: Solve for bb

Substitute a=6a = 6 into Equation 1: 64(6)+8b=43264(6) + 8b = 432 384+8b=432384 + 8b = 432 8b=48b=68b = 48 \quad \Rightarrow \quad b = 6

Final Answer:

The values of aa and bb that make the function continuous everywhere are: a=6,b=6a = 6, \quad b = 6

Would you like further details or have any questions?


Here are some related questions you might explore:

  1. What does it mean for a function to be continuous?
  2. How do you solve systems of linear equations in two variables?
  3. What are other conditions besides continuity that can apply to functions?
  4. How would you check differentiability at the points where the pieces of the function meet?
  5. What happens to the function if aa and bb take on different values?

Tip: When solving for continuity in piecewise functions, always check the behavior of the function at the points where different expressions are joined.

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Math Problem Analysis

Mathematical Concepts

Continuity
Piecewise Functions
Limits

Formulas

f(x) = (x^2 - 64) / (x + 8) for x < -8
f(x) = ax^2 - bx - 448 for -8 ≤ x < -1
f(x) = 436x - a + b for x ≥ -1

Theorems

Limit definition of continuity
Continuity at a point

Suitable Grade Level

Grade 11-12 (Pre-Calculus/Calculus)