Question Statement

Q14. Find the slope of the tangent to the curve $y=\frac{4x-1}{4}$ at $x=-1$.

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Background and Explanation

To find the slope of a tangent line to a curve at a specific point, we compute the derivative of the function and evaluate it at that $x$-value. This requires applying the quotient rule for differentiation if the function is expressed as a ratio of two expressions.

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Solution

Given the function: $y=\frac{4x-1}{4}$ and the point $x=-1$.

Differentiate $y$ with respect to $x$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{x\cdot \frac{d}{dx}(4x-1)-(4x-1)\frac{d}{dx}(x)}{x^2}\\ & =\frac{x\cdot (4-0)-(4x-1)(1)}{x^2}\\ & =\frac{4x-4x+1}{x^2}\\ \frac{dy}{dx}& =\frac{1}{x^2}\end{array}$

At $x=-1$:

$\frac{dy}{dx}=\frac{1}{(-1{)}^2}=\frac{1}{1}=1$

$\Rightarrow$ Slope of tangent $=1$

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Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$ (used to establish $\frac{d}{dx}(x)=1$)
• Constant Rule: $\frac{d}{dx}(c)=0$ (used to establish $\frac{d}{dx}(1)=0$)

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Summary of Steps

1. Identify the function $y=\frac{4x-1}{4}$ and the point of evaluation $x=-1$.
2. Apply the quotient rule to differentiate $y$ with respect to $x$.
3. Simplify the resulting expression to obtain $\frac{dy}{dx}=\frac{1}{x^2}$.
4. Substitute $x=-1$ into the derivative: $\frac{1}{(-1{)}^2}=1$.
5. Conclude that the slope of the tangent line at $x=-1$ is $1$.