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(c) \( h(x)=\frac{2}{x} \) \( \frac{h(x+h)}{h}=\frac{2}{x+h}= \)

Ask by Weaver Ayala. in the United States
Mar 11,2025

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Answer

\[ \frac{h(x+h)}{h} = -\frac{2}{x(x+h)} \]

Solution

We start with the function \[ h(x)=\frac{2}{x}. \] The difference quotient is defined by \[ \frac{h(x+h)-h(x)}{h}. \] Step 1: Compute \( h(x+h) \): \[ h(x+h)=\frac{2}{x+h}. \] Step 2: Write the expression for the difference quotient: \[ \frac{h(x+h)-h(x)}{h} = \frac{\frac{2}{x+h} - \frac{2}{x}}{h}. \] Step 3: Combine the fractions in the numerator: \[ \frac{2}{x+h} - \frac{2}{x} = \frac{2x - 2(x+h)}{x(x+h)}. \] Simplify the numerator: \[ 2x - 2(x+h) = 2x - 2x - 2h = -2h. \] So, the numerator becomes \[ \frac{-2h}{x(x+h)}. \] Step 4: Divide by \( h \): \[ \frac{\frac{-2h}{x(x+h)}}{h} = \frac{-2h}{h \cdot x(x+h)}. \] Cancel \( h \) (assuming \( h \neq 0 \)): \[ \frac{-2}{x(x+h)}. \] Thus, the final simplified expression is \[ -\frac{2}{x(x+h)}. \]

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The Deep Dive

When you substitute \( h \) into the function \( h(x+h) \), you essentially shift the input value of \( x \) to \( x + h \), which corresponds to the function output \( \frac{2}{x+h} \). Dividing by \( h \) keeps it as a fraction, allowing you to investigate the limit as \( h \) approaches zero. This is critical in calculus for finding derivatives, particularly the derivative of the function \( h(x) \). However, be careful not to make a common mistake here! Trying to evaluate directly by plugging in \( h = 0 \) can lead to undefined expressions. Instead, applying the limit process helps you deal effectively with those pesky \( h \)'s, leading you to the derivative of \( h(x) \) when examining \( \lim_{h \to 0} \frac{h(x+h) - h(x)}{h} \).

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