Question Statement

Find the slope of the tangent line to the curve $f(x)=x^3$ at the point $(1,f(1))$.

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Background and Explanation

To find the slope of a tangent line at a specific point, we use the limit definition of the derivative (the difference quotient). This method calculates the instantaneous rate of change by evaluating the limit of average rates of change as the interval approaches zero.

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Solution

Finding the Point of Tangency

First, evaluate the function at $x=1$ to determine the coordinates of the point:

$f(1)=(1{)}^3=1$

Therefore, the point of tangency is $(1,1)$.

Setting Up the Limit Definition

The slope of the tangent line is given by the derivative formula:

$m_{\tan }={\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

Given $f(x)=x^3$, we have:

$f(x+\Delta x)=(x+\Delta x{)}^3$

Expanding and Simplifying

Substitute into the limit definition and expand $(x+\Delta x{)}^3$ using the binomial theorem:

$\begin{array}{rl}{m}_{\tan }& =\underset{\Delta x\to 0}{\lim }\frac{(x+\Delta x{)}^3-x^3}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\frac{x^3+(\Delta x{)}^3+3x^2\Delta x+3x(\Delta x{)}^2-x^3}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\frac{(\Delta x{)}^3+3x^2\Delta x+3x(\Delta x{)}^2}{\Delta x}\end{array}$

Factor out $\Delta x$ from the numerator:

$\begin{array}{rl}& =\underset{\Delta x\to 0}{\lim }\frac{\Delta x\left((\Delta x{)}^2+3x^2+3x\Delta x\right)}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\left((\Delta x{)}^2+3x^2+3x\Delta x\right)\end{array}$

Evaluating the Limit

As $\Delta x\to 0$, the terms containing $\Delta x$ vanish:

$\begin{array}{rl}& =0+3x^2+0\\ & =3x^2\end{array}$

Thus, the slope of the tangent line at any point $x$ is:

$\text{Slope}=m_{\tan }=3x^2$

Finding the Slope at $x=1$

Substitute $x=1$ into the derivative:

$\begin{array}{rl}{m}_{\tan }& =3(1{)}^2\\ & =3(1)\\ & =3\end{array}$

Therefore, the slope of the tangent line at the point $(1,1)$ is 3.

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Key Formulas or Methods Used

• Limit Definition of the Derivative: $f^{\prime }(x)={\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$
• Binomial Expansion: $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$
• Slope Evaluation: Substitute the $x$-coordinate into the derivative $f^{\prime }(x)$ to find the slope at a specific point

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Summary of Steps

1. Identify the point: Calculate $f(1)=1$ to confirm the point is $(1,1)$
2. Apply the limit definition: Set up $m_{\tan }={\lim }_{\Delta x\to 0}\frac{(x+\Delta x{)}^3-x^3}{\Delta x}$
3. Expand the cubic: Use $(x+\Delta x{)}^3=x^3+3x^2\Delta x+3x(\Delta x{)}^2+(\Delta x{)}^3$
4. Simplify: Cancel $x^3$ terms, factor out $\Delta x$, and cancel with denominator
5. Evaluate the limit: As $\Delta x\to 0$, obtain $m_{\tan }=3x^2$
6. Calculate final slope: Substitute $x=1$ to get $m=3$