Problem 13 - Calculus Early Transcendentals

Question

Equations of tangent lines by definition (1) a. Use definition (1) (p. 127 ) to find the slope of the line tangent to the graph of \(f\) at \(P\) b. Determine an equation of the tangent line at \(P\). c. Plot the graph of \(f\) and the tangent line at \(P\). $$f(x)=\frac{1}{x} ; P(-1,-1)$$

Step-by-Step Solution

Answer

Question: Determine the equation of the tangent line to the function \(f(x) = \frac{1}{x}\) at the point \(P(-1,-1)\) and describe its graph. Answer: The equation of the tangent line to the function \(f(x) = \frac{1}{x}\) at the point \(P(-1,-1)\) is \(y = -x - 2\). The graph of the tangent line touches the curve of the function at point \(P(-1, -1)\), and it has a slope of -1 at that point.

Step 1: Find the derivative of f(x)

To find the tangent line's slope, we need to find the derivative of the given function \(f(x) = \frac{1}{x}\). To do this, use the power rule: $$ f'(x) = -\frac{1}{x^2} $$

Step 2: Evaluate the derivative at P

Now that we have the derivative, we can find the slope of the tangent line at the point \(P(-1,-1)\). Plugging \(x=-1\) into the derivative gives us: $$ f'(-1) = -\frac{1}{(-1)^2} = -1 $$ The slope of the tangent line at the point \(P(-1, -1)\) is \(-1\).

Step 3: Find the equation of the tangent line

We have the slope and the point \(P(-1,-1)\). We can use the point-slope form to find the equation of the tangent line: $$ y - y_1 = m(x - x_1) $$ Plugging in the values, we have: $$ y - (-1) = -1(x - (-1)) $$ Simplifying gives us the equation of the tangent line: $$ y + 1 = -x - 1 $$ $$ y = -x - 2 $$

Step 4: Plot the graph of f(x) and the tangent line

Plot the graph of the function \(f(x) = \frac{1}{x}\) and the tangent line \(y = -x - 2\) at the point \(P(-1, -1)\). The student should see the tangent line touch the curve at point P, and it should have a slope of -1 at that point.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at just one point without crossing it. Imagine a curve on a graph; the tangent line can be thought of as gently "kissing" the curve at a single point. At this point of contact, the curve and the line have the same slope. This unique feature makes the tangent line very useful in calculus, as it provides an approximation of the curve's behavior close to that point.

In the example given, the tangent line at point \(P(-1, -1)\) on the function \(f(x) = \frac{1}{x}\) touches the curve exactly at P. The slope of the tangent line is the same as the slope of the curve at this same point, which we've found to be -1. This means the tangent line has the equation \(y = -x - 2\). You could picture it as just lightly grazing the curve without cutting through it.

Derivative

The derivative of a function is a fundamental concept in calculus that measures how a function's output changes as its input changes. Simply put, it's like finding the slope of the function at any given point. When you're asked to find a derivative, you should think about finding the "instantaneous rate of change." For example, how fast is a car going at an exact moment?

For the function \(f(x) = \frac{1}{x}\), its derivative helps us determine the slope of the tangent line to the curve. Using calculus rules, specifically the power rule, we find the derivative is \(f'(x) = -\frac{1}{x^2}\). This derivative gives us the slope of the curve at any point \(x\). In our example, by plugging \(x = -1\) into the derivative, we compute \(f'(-1) = -1\). This is the slope of the tangent line at the point \(P\).

Power Rule

The power rule is a quick method in calculus for finding the derivative of expressions of the form \(x^n\), where \(n\) is any real number. The rule states that if you have a term \(x^n\), its derivative will be \(nx^{n-1}\). This makes it extremely handy for calculating derivatives of polynomials and functions that can be rewritten as polynomials.

For example, if we want to differentiate \(f(x) = \frac{1}{x}\), we can rewrite this as \(f(x) = x^{-1}\). Applying the power rule, the derivative is \(f'(x) = -1 \cdot x^{-2}\), which simplifies to \(-\frac{1}{x^2}\). It's a straightforward and efficient way to get to our answer, showing that the power rule can be a powerful tool in solving calculus problems.

Point-Slope Form

The point-slope form is a specific formula used in algebra to describe the equation of a line. It is particularly useful when you know one point on the line and the slope. The formula looks like this: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.

In the exercise, after finding that the slope \(m\) is -1 and knowing the point \(P(-1, -1)\), we used the point-slope form to find the tangent line equation. Plugging in the values gives us: \(y + 1 = -1(x + 1)\). Simplifying it, we obtain \(y = -x - 2\). This shows how using the point-slope form can straightforwardly convert slope information and a known point into the complete equation of a line.