Question Statement

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

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Background and Explanation

This problem requires using the limit definition of the derivative to find the slope of the tangent line at a specific point on a polynomial curve. The limit definition allows us to calculate the instantaneous rate of change by considering the limit of average rates of change over increasingly small intervals.

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Solution

We begin with the function: $f(x)=x^2-5x+4$

First, we determine $f(x+\Delta x)$ by substituting $(x+\Delta x)$ for every $x$ in the original function: $f(x+\Delta x)=(x+\Delta x{)}^2-5(x+\Delta x)+4$

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

Substituting our expressions for $f(x+\Delta x)$ and $f(x)$: $m_{\tan }={\lim }_{\Delta x\to 0}\frac{\left[(x+\Delta x{)}^2-5(x+\Delta x)+4\right]-\left[x^2-5x+4\right]}{\Delta x}$

Expanding the numerator: $(x+\Delta x{)}^2=x^2+2x\Delta x+(\Delta x{)}^2$ $-5(x+\Delta x)=-5x-5\Delta x$

So the numerator becomes: $\left[x^2+(\Delta x{)}^2+2x\Delta x-5x-5\Delta x+4\right]-\left[x^2-5x+4\right]$

Simplifying by distributing the negative sign and combining like terms: $=x^2+(\Delta x{)}^2+2x\Delta x-5x-5\Delta x+4-x^2+5x-4$ $=(\Delta x{)}^2+2x\Delta x-5\Delta x$

Now we substitute back into the limit: $m_{\tan }={\lim }_{\Delta x\to 0}\frac{(\Delta x{)}^2+2x\Delta x-5\Delta x}{\Delta x}$

Factor out $\Delta x$ from the numerator: $m_{\tan }={\lim }_{\Delta x\to 0}\frac{\Delta x(\Delta x+2x-5)}{\Delta x}$

Cancel $\Delta x$ from the numerator and denominator: $m_{\tan }={\lim }_{\Delta x\to 0}(\Delta x+2x-5)$

Evaluate the limit as $\Delta x\to 0$: $m_{\tan }=0+2x-5=2x-5$

Thus, the slope of the tangent line at any point $x$ is given by: $\text{Slope}=m_{\tan }=2x-5$

To find the slope at the specific point $(2,-2)$, we substitute $x=2$: $\begin{array}{rl}{m}_{\tan }& =2(2)-5\\ & =4-5\\ & =-1\end{array}$

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

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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: $m_{\tan }=f^{\prime }(x)$ evaluated at the specific $x$-coordinate
• Algebraic expansion: $(a+b{)}^2=a^2+2ab+b^2$

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Summary of Steps

1. Write the expression for $f(x+\Delta x)$ by substituting $(x+\Delta x)$ into the original function
2. Set up the limit definition of the derivative: ${\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$
3. Expand and simplify the numerator, combining like 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 $f^{\prime }(x)=2x-5$
6. Substitute $x=2$ (from the given point $(2,-2)$) into the derivative to find the specific slope $m=-1$