Problem 83 - Calculus and its applications

Question

For the function in each of graph \(\mathrm{f}\) and \(\mathrm{f}\). Then estimate points at which the tangent line to \(\mathrm{f}\) is horizontal. If no such point exists, state that fact. \(f(x)=\frac{4 x}{x^{2}+1}\)

Step-by-Step Solution

Answer

The tangent line is horizontal at \(x = 1\) and \(x = -1\).

Step 1: Identify When Tangent is Horizontal

A tangent line is horizontal at a point where the derivative of the function, \(f'(x)\), is equal to zero. Therefore, we need to find the derivative of the function and set it to zero.

Step 2: Find the Derivative of \(f(x)\)

The function is \(f(x) = \frac{4x}{x^2 + 1}\). We will use the quotient rule for this, which states: if \(h(x) = \frac{u(x)}{v(x)}\), then \(h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). Here, \(u(x) = 4x\) and \(v(x) = x^2 + 1\).

Step 3: Apply the Quotient Rule

Compute the derivatives: \(u'(x) = 4\) and \(v'(x) = 2x\). Substitute these into the quotient rule: \[ f'(x) = \frac{(4)(x^2 + 1) - (4x)(2x)}{(x^2 + 1)^2} \] Simplify the expression to find: \[f'(x) = \frac{4x^2 + 4 - 8x^2}{(x^2 + 1)^2} = \frac{-4x^2 + 4}{(x^2 + 1)^2}\].

Step 4: Solve \(f'(x) = 0\)

Set \(f'(x) = 0\): \[\frac{-4x^2 + 4}{(x^2 + 1)^2} = 0\]. The numerator must be zero, so solve: \[-4x^2 + 4 = 0\] This simplifies to: \[-4x^2 = -4\] Which leads to \[x^2 = 1\].

Step 5: Find the Points Where Tangent is Horizontal

Solve \(x^2 = 1\) to find the potential points: \(x = 1\) and \(x = -1\). These are the \(x\)-values where the tangent to the curve is horizontal.

Key Concepts

Quotient Rule

To calculate derivatives when dealing with a function that is a ratio of two expressions, we employ the Quotient Rule. This rule is vital in calculus as it helps us differentiate functions that are fractions.

The Quotient Rule states: if you have a function defined as a quotient of two functions, like:

• \( h(x) = \frac{u(x)}{v(x)}\)

Then, its derivative \( h'(x) \), is given by:

• \[ h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

To apply the rule, you need:

• The derivative of the numerator function \( u(x) \), which we denote as \( u'(x) \).
• The derivative of the denominator function \( v(x) \), denoted by \( v'(x) \).
• Then, plug into the formula above to get the derivative \( h'(x) \).

In our case with the function \( f(x) = \frac{4x}{x^2 + 1} \), we have:

• \( u(x) = 4x \) and consequently, \( u'(x) = 4 \).
• \( v(x) = x^2 + 1 \) with its derivative being \( v'(x) = 2x \).
• Substituting into the quotient rule yields the derivative \( f'(x) = \frac{4(x^2 + 1) - 4x \, (2x)}{(x^2 + 1)^2} \).

This rule simplifies the process of finding the derivative for complicated fraction functions.

Horizontal Tangent Line

A horizontal tangent line occurs at points on a graph where the slope is zero. This is significant in calculus because these points often indicate local maxima or minima. To find these horizontal tangent lines, we set the derivative of the function to zero.

For example, given a function \( f(x) \), the slope of its tangent line at any point is represented by \( f'(x) \). Therefore, horizontal tangents are where:

• \( f'(x) = 0 \)

In the case of our function:

• The derivative \( f'(x) = \frac{-4x^2 + 4}{(x^2 + 1)^2} \).
• To find the horizontal tangent lines, we solve: \[ -4x^2 + 4 = 0 \]
• This leads to \( x^2 = 1 \), giving the points \( x = 1 \) and \( x = -1 \) where the tangent is horizontal.

These solutions imply that at \( x = 1 \) and \( x = -1 \), the slope is zero, and therefore, the tangent lines are horizontal.

Derivative Calculation

The process of calculating derivatives is essential in understanding the behavior of a function. Derivatives tell us about rates of change and the slope of function graphs at any point.

For our function \( f(x) = \frac{4x}{x^2 + 1} \), we sought the derivative using the quotient rule. The steps included differentiation of both the numerator and the denominator separately.

• We found \( u'(x) = 4 \) and \( v'(x) = 2x \).
• Substituting these into the quotient rule yielded the expression \[ f'(x) = \frac{4(x^2 + 1) - 8x^2}{(x^2 + 1)^2} \].
• Simplifying gave us \( f'(x) = \frac{-4x^2 + 4}{(x^2 + 1)^2} \).

This expression represents the rate of change of \( f(x) \) at any given point. When this derivative is zero, it identifies potential horizontal tangents on the curve of the function.

Derivative calculations are tools that highlight how functions behave, enabling us to solve problems like finding when tangents are horizontal.