Problem 76 - A Graphical Approach to Precalculus with Limits

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) j for each function \(f\). Simplify completely. $$f(x)=5 x-6$$

Step-by-Step Solution

Answer

The difference quotient simplifies to 5.

Step 1: Identify the expression for f(x+h)

First, we need to know what expression to plug into the formula. We start with the given function \( f(x) = 5x - 6 \). To find \( f(x+h) \), substitute \( x + h \) into the function instead of \( x \). This gives us \( f(x+h) = 5(x + h) - 6 \).

Step 2: Expand f(x+h)

Now, let's expand this expression: \( f(x+h) = 5(x+h) - 6 = 5x + 5h - 6 \). This represents the expression for \( f(x+h) \).

Step 3: Substitute into the Difference Quotient

The difference quotient is given by \( \frac{f(x+h) - f(x)}{h} \). Substitute the expressions for \( f(x+h) \) and \( f(x) \) into this quotient:\[\frac{(5x + 5h - 6) - (5x - 6)}{h}\]

Step 4: Simplify the Numerator

Simplify by distributing and combining like terms in the numerator:\[(5x + 5h - 6) - (5x - 6) = 5x + 5h - 6 - 5x + 6\]This simplifies to:\[5h\]

Step 5: Complete the Difference Quotient

The difference quotient now simplifies to:\[\frac{5h}{h}\]Since \( h eq 0 \), divide the \( h \) in the numerator and denominator to get:\[5\]

Step 6: Write the Final Result

The final result after simplification is that the difference quotient equals 5.

Key Concepts

Function Evaluation

Function evaluation involves finding the value of a function for a specific input. In this exercise, we start with a linear function, given as \( f(x) = 5x - 6 \). The task is to evaluate the function at \( x + h \), not just at \( x \).
To do this, substitute \( x + h \) into the function in place of \( x \). This gives us a new expression: \( f(x + h) = 5(x + h) - 6 \). The evaluation here involves replacing every \( x \) in the function with \( x + h \) and then simplifying.

• This substitution helps us to build the expression needed for the difference quotient.
• To evaluate the function correctly at \( x + h \), ensure every term involving \( x \) is adjusted to include \( h \).
• Accurate function evaluation is crucial for obtaining a correct difference quotient.

Simplification

Simplification is the process of rewriting an expression in its most concise form. It reduces the complexity of expressions either by expanding, factoring or canceling terms.
In our exercise, after finding \( f(x+h) = 5(x + h) - 6 \), expand it to \( 5x + 5h - 6 \). Simplification hasn't finished here, as it extends to more than just multiplying terms.
Next, substitute back into the difference quotient using the results from function evaluation:\[\frac{(5x + 5h - 6) - (5x - 6)}{h}\]

• The subtraction inside the bracket leads to \( 5x + 5h - 6 - 5x + 6 \).
• Combine like terms to simplify this expression to \( 5h \).
• Such simplification lets us see clearly that all terms apart from \( 5h \) cancel out, focusing squarely on the term involving \( h \).

Algebraic Manipulation

Algebraic manipulation is key in rearranging and simplifying expressions or equations to solve them efficiently. In the context of the difference quotient, this means transforming our expression using basic algebra rules.
At the last step of our computation, we have the expression \( \frac{5h}{h} \). Remember, \( h eq 0 \) as stated in the problem, or else the expression would be undefined.

• Cancel out \( h \) in the numerator and denominator, which simplifies the expression to 5.
• Make sure that any division by a variable assumes the variable isn't zero, emphasizing the importance of keeping \( h eq 0 \).
• This manipulation finalizes the process, showing us that the difference quotient equals 5.

Algebraic manipulation is vital for solving problems like these, as it strips down and reveals the underlying relations in mathematical expressions.