Math Problem Statement

Let f(x) be the function 1/(x + 8). Then the quotient (f(6 + h) - f(6)) / h can be simplified to -1 / (ah + b) for:

Solution

The problem provided involves analyzing the difference quotient for the function f(x)=1x+8f(x) = \frac{1}{x + 8}. We need to simplify the expression

f(6+h)f(6)h\frac{f(6 + h) - f(6)}{h}

and then express it in the form 1ah+b\frac{-1}{ah + b}.

Step-by-Step Solution:

  1. Substitute f(x)f(x) into the difference quotient:

    Since f(x)=1x+8f(x) = \frac{1}{x + 8}, we can compute f(6+h)f(6 + h) and f(6)f(6) as follows:

    • f(6+h)=1(6+h)+8=1h+14f(6 + h) = \frac{1}{(6 + h) + 8} = \frac{1}{h + 14}
    • f(6)=16+8=114f(6) = \frac{1}{6 + 8} = \frac{1}{14}

    So, the difference quotient becomes:

    f(6+h)f(6)h=1h+14114h\frac{f(6 + h) - f(6)}{h} = \frac{\frac{1}{h + 14} - \frac{1}{14}}{h}

  2. Simplify the numerator:

    Find a common denominator for 1h+14114\frac{1}{h + 14} - \frac{1}{14}:

    1h+14114=14(h+14)14(h+14)=14h1414(h+14)=h14(h+14)\frac{1}{h + 14} - \frac{1}{14} = \frac{14 - (h + 14)}{14(h + 14)} = \frac{14 - h - 14}{14(h + 14)} = \frac{-h}{14(h + 14)}

    So now we have:

    f(6+h)f(6)h=h14(h+14)h=h14(h+14)h\frac{f(6 + h) - f(6)}{h} = \frac{\frac{-h}{14(h + 14)}}{h} = \frac{-h}{14(h + 14) \cdot h}

  3. Cancel hh from the numerator and denominator:

    h14(h+14)h=114(h+14)\frac{-h}{14(h + 14) \cdot h} = \frac{-1}{14(h + 14)}

  4. Final Form:

    Now we have simplified the expression to:

    f(6+h)f(6)h=114(h+14)\frac{f(6 + h) - f(6)}{h} = \frac{-1}{14(h + 14)}

Answer:

The quotient can be simplified to:

1ah+bwhere a=14 and b=196.\frac{-1}{ah + b} \quad \text{where } a = 14 \text{ and } b = 196.

Would you like further details on this solution?


Related Questions:

  1. What is the difference quotient, and why is it used in calculus?
  2. How would the solution change if the function were f(x)=1x+cf(x) = \frac{1}{x + c} with a different constant cc?
  3. What is the significance of finding the limit as h0h \to 0 in the difference quotient?
  4. How does this result relate to finding the derivative of f(x)=1x+8f(x) = \frac{1}{x + 8}?
  5. What other techniques could be used to simplify complex rational expressions?

Tip:

In difference quotient problems, always simplify the numerator carefully by finding a common denominator to avoid errors in handling complex fractions.

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Math Problem Analysis

Mathematical Concepts

Difference Quotient
Function Evaluation
Algebraic Simplification

Formulas

Difference Quotient Formula: (f(x + h) - f(x)) / h

Theorems

Limit Definition of Derivative (conceptually related)

Suitable Grade Level

Grades 11-12