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Chapter 2: Problem 89

Find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for each function and simplify it. $$y=-x^{2}+x-2$$

Short Answer

Expert verified
\( -2x - h + 1 \).

Step by step solution

01

- Identify the function

The given function is \(f(x) = -x^2 + x - 2\).
02

- Find \(f(x+h)\)

Replace \(x\) with \(x+h\) in the function: \(f(x+h) = -(x+h)^2 + (x+h) - 2\).
03

- Expand \(f(x+h)\)

Expand the expression: \(f(x+h) = -(x^2 + 2xh + h^2) + x + h - 2\) Simplify it: \(f(x+h) = -x^2 - 2xh - h^2 + x + h - 2\).
04

- Substitute \(f(x)\) and \(f(x+h)\) into the difference quotient

The difference quotient is \( \frac{f(x+h) - f(x)}{h} \) Substitute \(f(x+h)\) and \(f(x)\): \( \frac{[-x^2 - 2xh - h^2 + x + h - 2] - [-x^2 + x - 2]}{h} \).
05

- Simplify the numerator

Combine like terms in the numerator: \( -x^2 - 2xh - h^2 + x + h - 2 + x^2 - x + 2 \) Which results in: \( -2xh - h^2 + h \).
06

- Final simplification

Factor out ~~\(h)\) in the numerator: \( \frac{h(-2x - h + 1)}{h} \) Cancel the \(h\) term: \( -2x - h + 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

precalculus
Precalculus builds a necessary foundation for calculus by covering essential algebraic and trigonometric concepts. One such concept is the difference quotient. The difference quotient is critical in calculus for calculating the derivative of a function, which measures how a function's output changes with respect to changes in its input.
function expansion
For the difference quotient, you need to compute values like \( f(x+h) \). This requires function expansion. For instance, with our function \( f(x) = -x^2 + x - 2 \), we substitute \( x + h \) into the function to get \( f(x+h) \). Thus, expanding \( -(x+h)^2 + (x+h) - 2 \) is essential:
  • First, square \(x + h\) to get \(x^2 + 2xh + h^2\).
  • Then distribute the negative sign: \(-(x^2 + 2xh + h^2) = -x^2 - 2xh - h^2\).
  • Add the linear terms \(x + h\) and subtract 2: \(-x^2 - 2xh - h^2 + x + h - 2\).
algebraic simplification
In algebraic simplification, you combine like terms and factor when possible. After expanding the function as in the previous step, plug values into the difference quotient: \( \frac{[-x^2 - 2xh - h^2 + x + h - 2] - [-x^2 + x - 2]}{h} \). Simplify the numerator by combining like terms:
  • Cancel out terms: \( -x^2 \) and \( x \).

  • This leaves \(-2xh - h^2 + h \).

Factor out \(h\) from the simplified numerator: \( \frac{h(-2x - h + 1)}{h} \). Finally, cancel the \(h\) terms: \( -2x - h + 1 \).
derivatives preparation
The difference quotient prepares you for calculus' concept of derivatives. By deriving the simplified form \( -2x - h + 1 \), you're essentially forming the basis of finding a function's rate of change. As \( h \) approaches zero, the simplified difference quotient \( -2x + 1 \) becomes the derivative \( \frac{d}{dx}(-x^2 + x - 2) \). Practicing these steps in precalculus enables a smooth transition to calculus, where you'll make use of these simplification techniques regularly to find instantaneous rates of change.

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