Question Statement

Find the slope of the tangent line to the function $f(x)=\frac{1}{x}$ at the point $\left(\frac{1}{3},f\left(\frac{1}{3}\right)\right)$.

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Background and Explanation

This problem requires calculating the instantaneous rate of change using the limit definition of the derivative. You will need to manipulate complex fractions algebraically to evaluate the limit, which yields the general derivative formula before substituting the specific $x$-value.

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Solution

First, determine the coordinates of the point of tangency by evaluating $f\left(\frac{1}{3}\right)$:

$f\left(\frac{1}{3}\right)=\frac{1}{1/3}=3$

$\therefore$ The point is $\left(\frac{1}{3},3\right)$.

Given the function definitions: $\begin{array}{rl}f(x)& =\frac{1}{x}\\ f(x+\Delta x)& =\frac{1}{x+\Delta x}\end{array}$

The slope of the tangent line is given by the limit definition of the derivative: $\text{Slope of tangent}=m_{\tan }={\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

Substituting the function into the limit expression: $\begin{array}{rl}& =\underset{\Delta x\to 0}{\lim }\frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\frac{1}{\Delta x}\left[\frac{1}{x+\Delta x}-\frac{1}{x}\right]\\ & =\underset{\Delta x\to 0}{\lim }\frac{1}{\Delta x}\left[\frac{x-1(x+\Delta x)}{(x+\Delta x)\cdot x}\right]\\ & =\underset{\Delta x\to 0}{\lim }\frac{1}{\Delta x}\left[\frac{x-x-\Delta x}{(x+\Delta x)\cdot x}\right]\\ & =\underset{\Delta x\to 0}{\lim }\frac{1}{\Delta x}\left[\frac{-\Delta x}{(x+\Delta x)\cdot x}\right]\\ & =\underset{\Delta x\to 0}{\lim }\frac{-1}{(x+\Delta x)\cdot x}\\ & =\frac{-1}{(x+0)\cdot x}\\ & =\frac{-1}{x^2}\end{array}$

Now evaluate the slope at the point where $x=\frac{1}{3}$: $\begin{array}{rl}\text{Slope}=m_{\tan }& =\frac{-1}{{\left(\frac{1}{3}\right)}^2}\\ & =\frac{-1}{\frac{1}{9}}\\ & =-9\end{array}$

Therefore, the slope of the tangent line is $m_{\tan }=-9$.

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Key Formulas or Methods Used

• Limit definition of the derivative: $f^{\prime }(x)={\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$
• Slope of tangent line: $m_{\tan }=f^{\prime }(a)$ at point $x=a$
• Algebraic simplification of complex fractions (finding common denominators and canceling terms)

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Summary of Steps

1. Evaluate $f\left(\frac{1}{3}\right)=3$ to confirm the point of tangency is $\left(\frac{1}{3},3\right)$.
2. Set up the limit definition of the derivative with $f(x)=\frac{1}{x}$ and $f(x+\Delta x)=\frac{1}{x+\Delta x}$.
3. Combine the fractions in the numerator by finding a common denominator $x(x+\Delta x)$.
4. Simplify the numerator to $-\Delta x$, cancel $\Delta x$ from numerator and denominator, and evaluate the limit as $\Delta x\to 0$ to obtain $f^{\prime }(x)=-\frac{1}{x^2}$.
5. Substitute $x=\frac{1}{3}$ into the derivative formula to calculate the specific slope $m_{\tan }=-9$.