Problem 65 - Applied Calculus

Question

For each function, find and simplify \(\frac{f(x+h)-f(x)}{h} .\) (Assume \(\left.h \neq 0 .\right)\) (See instructions on previous page.) $$ f(x)=2 x^{2}-5 x+1 $$

Step-by-Step Solution

Answer

\( 4x + 2h - 5 \)

Step 1: Express f(x+h)

First, express the function \( f(x+h) \) by substituting \( x+h \) into the original function \( f(x) = 2x^2 - 5x + 1 \). So, \[ f(x+h) = 2(x+h)^2 - 5(x+h) + 1. \]

Step 2: Expand f(x+h)

Next, expand the terms in \( f(x+h) \). Begin with \( 2(x+h)^2 \) which becomes \( 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2 \). Then, expand \( -5(x+h) \) to obtain \( -5x - 5h \). Putting it all together, we get: \[ f(x+h) = 2x^2 + 4xh + 2h^2 - 5x - 5h + 1. \]

Step 3: Formulate the Difference Quotient

With \( f(x+h) \) and \( f(x) \) known, form the difference quotient \( \frac{f(x+h) - f(x)}{h} \). Subtract \( f(x) = 2x^2 - 5x + 1 \) from \( f(x+h) = 2x^2 + 4xh + 2h^2 - 5x - 5h + 1 \), which simplifies to: \[ f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 5x - 5h + 1) - (2x^2 - 5x + 1). \] Simplify the expression: \( 4xh + 2h^2 - 5h \).

Step 4: Simplify the Expression

The difference becomes \( 4xh + 2h^2 - 5h \). Factor \( h \) out of the expression: \[ f(x+h) - f(x) = h(4x + 2h - 5). \] Now divide each term by \( h \) (since \( h eq 0 \)): \[ \frac{f(x+h) - f(x)}{h} = \frac{h(4x + 2h - 5)}{h} = 4x + 2h - 5. \]

Step 5: Final Result

Thus, the simplified expression for the difference quotient is \( 4x + 2h - 5 \). Make sure to include any restrictions or conditions specified; here \( h eq 0 \).

Key Concepts

Function Expansion

To fully understand how to form the difference quotient, it starts with function expansion. When you have a function like \( f(x) = 2x^2 - 5x + 1 \), to find \( f(x+h) \), substitute \( x+h \) in place of \( x \). This gives you \( f(x+h) = 2(x+h)^2 - 5(x+h) + 1 \). Expanding involves distributing and simplifying each component:

• Begin with \( 2(x+h)^2 \): Use the formula \((a+b)^2 = a^2 + 2ab + b^2\) to get \( 2x^2 + 4xh + 2h^2 \).
• For \( -5(x+h) \), distribute \(-5\) to both \( x \) and \( h \), resulting in \(-5x - 5h\).
• Combine all pieces: \( 2x^2 + 4xh + 2h^2 - 5x - 5h + 1\).

Expanding functions is crucial for breaking down more complicated expressions into manageable parts. By getting \( f(x+h) \) in its expanded form, further simplification becomes possible.

Simplifying Expressions

Once you have expanded \( f(x+h) \), the next critical step is simplifying expressions. For the given problem, you need the expression \( f(x+h) - f(x) \). Subtract the original function \( f(x) = 2x^2 - 5x + 1 \) from the expanded form:

• Subtract \( 2x^2 - 5x + 1 \) from \( 2x^2 + 4xh + 2h^2 - 5x - 5h + 1 \).
• On cancellation, terms like \( 2x^2 \), \(-5x\), and \(+1\) cancel out respectively.
• This leads to: \( 4xh + 2h^2 - 5h \).

Simplifying here involves factoring and canceling terms where possible. In this case, factoring \( h \) out of the expression as \( h(4x + 2h - 5) \) reveals the core components of the quotient. By understanding simplification, you make complex equations easier to manage, setting the stage for the final step.

Calculus Problem Solving

Calculus problem solving often necessitates developing and manipulating difference quotients. The purpose of the difference quotient, \( \frac{f(x+h) - f(x)}{h} \), lies in understanding changes in a function.

• After simplifying the expression to \( h(4x + 2h - 5) \), divide each term by \( h \), crucially noting \( h eq 0 \).
• This results in \( 4x + 2h - 5 \), removing \( h \) through the cancellation of factors.

Successfully solving calculus problems involves logic and recognizing patterns in algebraic manipulation. The process highlights how calculus builds upon foundational algebra skills—like factoring and simplifying expressions—to provide deeper insights into how functions behave.