Problem 50 - Calculus Early Transcendentals

Question

Use a graphing utility to make rough estimates of the locations of all horizontal tangent lines, and then find their exact locations by differentiating. $$ y=\frac{x^{2}+9}{x} $$

Step-by-Step Solution

Answer

Horizontal tangents are at \((3, 6)\) and \((-3, -6)\).

Step 1: Rewrite the function

Rewrite the given function in a format suitable for differentiation. The function is given as \( y = \frac{x^2 + 9}{x} \). This can be simplified to \( y = x + \frac{9}{x} \) since dividing each term by \( x \) yields this form.

Step 2: Differentiate the function

Differentiate the simplified function using the power rule. The function is \( y = x + \frac{9}{x} \). The derivative is \( \frac{dy}{dx} = 1 - \frac{9}{x^2} \).

Step 3: Set derivative to zero for horizontal tangents

Horizontal tangent lines occur where the derivative equals zero. Set \( \frac{dy}{dx} = 0 \):\[ 1 - \frac{9}{x^2} = 0 \] Solve for \( x \) which gives \( x^2 = 9 \).

Step 4: Solve for x

Take the square root of both sides of \( x^2 = 9 \) to find \( x = 3 \) or \( x = -3 \). These are the x-coordinates where the horizontal tangents occur.

Step 5: Find the y-coordinates

Substitute \( x = 3 \) and \( x = -3 \) into the original function to find the corresponding y-values. For \( x = 3 \):\[ y = \frac{3^2 + 9}{3} = \frac{18}{3} = 6 \] For \( x = -3 \):\[ y = \frac{(-3)^2 + 9}{-3} = \frac{18}{-3} = -6 \].Thus, the points are \( (3, 6) \) and \( (-3, -6) \).

Key Concepts

Differentiation

Differentiation is a cornerstone concept in calculus. It helps us understand how a function changes. It tells us rates of change and gives us the slope of a curve at any given point. When we differentiate a function, we find its derivative. The derivative is a new function which gives the slope of the original function at any x-value.

For example, if you think of a curve as a hilly road, differentiation lets us know how steep or flat the road is at every point. This is especially useful when trying to understand the behavior of curves in varying situations like physics or economics.

In our problem, we started with the function

• \( y = \frac{x^2 + 9}{x} \)

By simplifying it to

• \( y = x + \frac{9}{x} \),

we made it easier to find the derivative. A derivative gives us all the information about the slope and how the function behaves.

Horizontal Tangents

Horizontal tangents are special points on a curve where the slope is zero. This means the curve flattens out at those points. When a car is driving on a hill, a horizontal tangent is like reaching the top and driving on a flat section before starting to go downhill again.

To find where these occur, we look for when the derivative equals zero. This is because a zero slope means a flat tangent line. In our problem, once we differentiated, we set the derivative expression

• \( \frac{dy}{dx} = 1 - \frac{9}{x^2} \)

to zero:

\[ 1 - \frac{9}{x^2} = 0 \].

Solving this gives us the x-values where horizontal tangents exist. Here, the solutions are \( x = 3 \) and \( x = -3 \). These points tell us where the curve flattens out, revealing key characteristics of the curve's behavior at these precise spots.

Power Rule

The power rule is a simple formula used to find derivatives of functions of the form \( x^n \), where \( n \) is any real number. It is one of the first rules in differentiation and is extremely useful for breaking down more complex expressions.

The power rule states: If \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \). This rule helps us to quickly and easily find the slope of polynomial terms.

In our problem, we used the power rule to differentiate the simplified function

• \( y = x + \frac{9}{x} \),

which can be seen as

• \( y = x^1 + 9x^{-1} \).

Applying the power rule gives

• \( \frac{dy}{dx} = 1 \times x^{0} - 9 \times x^{-2} \),

which simplifies to

• \( \frac{dy}{dx} = 1 - \frac{9}{x^2} \).

This approach shows how straightforward it can be to differentiate polynomial terms using the power rule, making it indispensable for solving problems like these.