Problem 31 - Calculus of a Single Variable

Question

Finding an Equation of a Tangent Line In Exercises \(25-32,(\text { a) find an equation of the tangent line to the graph of } f\) at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$ f(x)=x+\frac{4}{x}, \quad(-4,-5) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(f(x)=x+\frac{4}{x}\) at the point (-4,-5) is \(y=0.75x+2\).

Step 1: Find the Derivative of the Function

The derivative of the function can be found using the power and quotient rules. The derivative of \(f(x)=x+\frac{4}{x}\) is \(f'(x)=1-\frac{4}{x^2}\).

Step 2: Plug in x-value of the Given Point to Find the Slope

The slope of the tangent line can be found by evaluating the derivative at the x-value of the given point. Plugging in -4 into the derivative gives \(f'(-4)=1-\frac{4}{(-4)^2}=1-\frac{1}{4}=0.75\). Therefore, the slope of the tangent line is 0.75.

Step 3: Use Point-Slope Form to Find the Equation of the Tangent Line

The point-slope form of a line is \(y-y1=m(x-x1)\), where m is the slope and \((x1,y1)\) is a point on the line. Using the given point (-4,-5) and the slope 0.75, the equation of the tangent line is \(y-(-5)=0.75(x-(-4))\). Simplifying this gives the equation \(y=0.75x+2\).

Step 4: Plot the function and the found tangent line

Plot the function \(f(x)=x+\frac{4}{x}\) and the tangent line \(y=0.75x+2\).

Step 5: Validate results

Using the derivative feature of the graphing utility, confirm that the derived results are accurate.

Step 6: Wrap up

Verify that the drawn line is indeed tangent to the original function at the given point (-4,-5), and that the computed derivative matches the derivative from the graphing utility.

Key Concepts

Derivative of a Function

Understanding the derivative of a function is crucial when finding the equation of a tangent line. The derivative represents the rate of change or the slope of the function at any given point. Imagine you're driving and the steepness of the road changes as you move – the derivative would tell you how steep the road is at any moment.

For the function given in our exercise, f(x) = x + \(\frac{4}{x}\), the derivative f'(x) reflects how quickly f(x) is changing as x changes. Calculating the derivative allows us to determine the exact slope of the tangent line at a specific point on the function's graph. This slope can be thought of as how tilted the line is at that very spot. Without finding the derivative, we wouldn't have a precise value for the tangent's slope and couldn't accurately draw the line.

Power and Quotient Rules

To solve our textbook problem, we need to employ both the power and quotient rules of differentiation. These handy rules help us to find derivatives of functions involving powers of x, or fractions involving x.

The power rule tells us that to differentiate xⁿ, we multiply by the exponent n, then subtract one from the exponent. For instance, the derivative of x³ is 3x². Simple, isn't it?

When we have a fraction, like \(\frac{4}{x}\), the quotient rule can be bypassed in favor of the power rule by writing the denominator as x^-1. We did this for our function's derivative, f'(x) = 1 - \(\frac{4}{x^2}\), which combines the derivatives of x and \(\frac{4}{x}\), making our work much simpler.

Point-Slope Form

With our slope in hand from the derivative, it's time to frame the equation of our tangent line using the point-slope form. This is a straightforward way to write the equation of a line when we know a point it passes through and its slope.

The general equation for the point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is the point on the line, and m is the slope. For our problem, we fill in the point (-4, -5) and slope 0.75 into the formula, creating an equation that uniquely defines the tangent line at that point on the graph of f(x). It's like giving you the exact direction and location to draw a straight line on a map that just grazes a curve at a single point - not going through it, not moving away from it, but touching it precisely at one spot.