Problem 21 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the slope of the tangent line to the graph of each function at the given point and determine an equation of the tangent line. \(f(x)=-\frac{1}{x}\) at \(\left(3,-\frac{1}{3}\right)\)

Step-by-Step Solution

Answer

To find the equation of the tangent line to the graph of the function \(f(x)=-\frac{1}{x}\) at the point \(\left(3,-\frac{1}{3}\right)\), first differentiate the function with respect to x: \(f'(x) = x^{-2}\). Next, evaluate the derivative at the given x-value: \(f'(3) = \frac{1}{9}\). Then, apply point-slope form with the slope and given point, yielding the final equation of the tangent line: \(y = \frac{1}{9}x - \frac{2}{3}\).

Step 1: Differentiate the function with respect to x

Differentiate the function: \(f(x) = -\frac{1}{x}\) Find the derivative: \(f'(x) = \frac{d}{dx} \left(-\frac{1}{x}\right)\) To do this, we can rewrite the function as: \(f(x) = -x^{-1}\) Now we can differentiate using the power rule: \(f'(x) = -(-1)x^{-2} = x^{-2}\) So, \(f'(x) = x^{-2}\)

Step 2: Evaluate the derivative at the x-value of the given point

Now, evaluate the derivative at the given x-value: \(x = 3\) \(f'(3) = 3^{-2} = \frac{1}{9}\) So the slope of the tangent line at the given x-value: \(m = \frac{1}{9}\)

Step 3: Use the point-slope form to find an equation of the tangent line

Use the point-slope form of the linear equation: \(y - y_1 = m(x - x_1)\) Plug in the values from the given point \((3, -\frac{1}{3})\) and the slope we found, \(\frac{1}{9}\). \(y + \frac{1}{3} = \frac{1}{9}(x - 3)\) Now, we can solve for y to find the equation of the tangent line. Multiply both sides by 9 to eliminate the fractions: \(9y + 3 = 1(x - 3)\) Distribute and simplify: \(9y + 3 = x - 3\) \(9y = x - 6\) Now we have the equation of the tangent line: \(\boxed{y = \frac{1}{9}x - \frac{2}{3}}\)

Key Concepts

Tangent Line

The concept of a tangent line is central in calculus for understanding curves. A tangent line touches a curve at exactly one point, without crossing it. This means the tangent line "grazes" the curve at that point. In simpler terms, if you imagine a curve as a hill, the tangent line would be like the path you'd walk if you were trying to stay level at just that one point. This line shows the direction the curve is heading at that particular spot. Looking at the point of tangency provides valuable insights into the behavior of the curve at that location.

Derivative

Derivatives are a fundamental concept in calculus, helping us find the rate at which things change. When we talk about the derivative of a function, we're discussing how the function's output changes as its input changes. In our case, the function is given by \(f(x) = -\frac{1}{x}\). By differentiating this function, we determine how the slope of the tangent line will be at any point \(x\). Differentiation might seem complex, but a simple rule often helps: the power rule. For the function at hand, expressing it as \(-x^{-1}\) makes it easy to apply this rule to find the derivative \(f'(x) = x^{-2}\). This derivative is crucial because it gives us the slope of the tangent line, an essential component in sketching the line.

Point-Slope Form

The point-slope form is an equation type that simplifies finding the equation of a line if we know the slope and a point on the line. It looks like this: \(y - y_1 = m(x - x_1)\). Here, \(m\) represents the slope, while \((x_1, y_1)\) is a point through which the line passes. It's perfect for our tangent line problem because we already have the slope at \(x = 3\) and the point \((3, -\frac{1}{3})\). Using this format helps us quickly write the equation of our tangent line.

• They are efficient in drafting the equation.
• They connect to the core concept of slope, highlighted by the derivative.
• They simplify the algebra needed to form equations.

By inserting our known values into the point-slope formula, we derive an equation representing the tangent exactly where the curve touches the line.

Slope of Tangent Line

The slope of a tangent line is a numerical representation of its steepness and direction. At any point on a curve, it shows us the immediate rate of change.

• If the slope is positive, the tangent line rises; if negative, it falls.
• A zero slope indicates a perfectly flat, horizontal line.
• The slope is calculated by evaluating the derivative at the specific x-value of interest.

In our exercise, finding the derivative \(f'(x) = x^{-2}\) allowed us to determine the slope at \(x = 3\), namely \(\frac{1}{9}\). This small positive number shows a gentle upward rise at the tangent point \((3, -\frac{1}{3})\). Understanding this slope helps not only in visualizing how the curve behaves but also in forming the precise equation of the tangent line. This understanding is pivotal in crafting the exact line that just barely kisses the curve at the desired location.