Question Statement

Find the slope of the tangent line to the curve $f(x)=x^2+4$ at the point $(-1,5)$.

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Background and Explanation

This problem requires finding the derivative of a function using the limit definition (the difference quotient), which represents the slope of the tangent line at any point $x$. The derivative measures the instantaneous rate of change of the function at a specific point.

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Solution

We begin with the given function: $f(x)=x^2+4$

First, we determine $f(x+\Delta x)$ by substituting $(x+\Delta x)$ into the function: $f(x+\Delta x)=(x+\Delta x{)}^2+4$

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

Substituting the expressions for $f(x+\Delta x)$ and $f(x)$ into the limit definition:

$\begin{array}{rl}{m}_{\tan }& =\underset{\Delta x\to 0}{\lim }\frac{\left((x+\Delta x{)}^2+4\right)-\left(x^2+4\right)}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\frac{x^2+(\Delta x{)}^2+2x\Delta x+4-x^2-4}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\frac{\Delta x(\Delta x+2x)}{\Delta x}\\ & =0+2x\end{array}$

Thus, the slope of the tangent line at any point $x$ is: $\text{Slope}=m_{\tan }=2x$

At the point $(-1,5)$, we substitute $x=-1$: $\text{Slope}=m_{\tan }=2(-1)=-2$

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Key Formulas or Methods Used

• Limit definition of the derivative (slope of tangent): $m_{\tan }={\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$
• Binomial expansion: $(x+\Delta x{)}^2=x^2+2x\Delta x+(\Delta x{)}^2$
• Limit evaluation: ${\lim }_{\Delta x\to 0}(\Delta x+2x)=2x$

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Summary of Steps

1. Write $f(x)=x^2+4$ and find $f(x+\Delta x)=(x+\Delta x{)}^2+4$
2. Set up the difference quotient: ${\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$
3. Expand $(x+\Delta x{)}^2$ and simplify the numerator by canceling $x^2$ and constant terms
4. Factor out $\Delta x$ from the numerator and cancel with the denominator
5. Evaluate the limit as $\Delta x\to 0$ to obtain the derivative $2x$
6. Substitute $x=-1$ to find the specific slope: $-2$