Simplify the difference quotients ƒ(x+h) - ƒ(x) / h and ƒ(x) - ƒ(a) / (x-a) by rationalizing the numerator.
ƒ(x) = √(1-2x)

검증된 단계별 안내

First, identify the function given: \( f(x) = \sqrt{1-2x} \). We need to simplify the difference quotients \( \frac{f(x+h) - f(x)}{h} \) and \( \frac{f(x) - f(a)}{x-a} \).

For the first difference quotient \( \frac{f(x+h) - f(x)}{h} \), substitute \( f(x+h) = \sqrt{1-2(x+h)} = \sqrt{1-2x-2h} \). The expression becomes \( \frac{\sqrt{1-2x-2h} - \sqrt{1-2x}}{h} \).

To rationalize the numerator, multiply the numerator and the denominator by the conjugate of the numerator: \( \frac{\sqrt{1-2x-2h} - \sqrt{1-2x}}{h} \times \frac{\sqrt{1-2x-2h} + \sqrt{1-2x}}{\sqrt{1-2x-2h} + \sqrt{1-2x}} \).

Simplify the numerator using the difference of squares: \( (\sqrt{1-2x-2h})^2 - (\sqrt{1-2x})^2 = (1-2x-2h) - (1-2x) = -2h \). The expression becomes \( \frac{-2h}{h(\sqrt{1-2x-2h} + \sqrt{1-2x})} \).

Cancel \( h \) in the numerator and denominator: \( \frac{-2}{\sqrt{1-2x-2h} + \sqrt{1-2x}} \). This is the simplified form of the first difference quotient. Repeat a similar process for the second difference quotient \( \frac{f(x) - f(a)}{x-a} \) by substituting \( f(a) = \sqrt{1-2a} \) and rationalizing the numerator.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

11m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Difference Quotient

The difference quotient is a fundamental concept in calculus that represents the average rate of change of a function over an interval. It is expressed as (ƒ(x+h) - ƒ(x)) / h, where h is the change in x. This concept is crucial for understanding derivatives, as the limit of the difference quotient as h approaches zero gives the derivative of the function.

Rationalizing the Numerator

Rationalizing the numerator involves manipulating an expression to eliminate any irrational numbers from the numerator. This is often done by multiplying the numerator and denominator by the conjugate of the numerator. This technique simplifies expressions, making it easier to evaluate limits or perform algebraic operations, especially in calculus.

Function Composition

Function composition is the process of applying one function to the results of another function. In the context of the given question, understanding how to evaluate ƒ(x+h) involves substituting (x+h) into the function ƒ(x) = √(1-2x). This concept is essential for simplifying expressions and understanding how changes in input affect the output of a function.

Simplify the difference quotients ƒ(x+h) - ƒ(x) / h and ƒ(x) - ƒ(a) / (x-a) by rationalizing the numerator.ƒ(x) = √(1-2x)