Problem 49 - College Algebra and Calculus: An Applied Approach

Question

Find the point(s), if any, at which the graph of \(f\) has a horizontal tangent. $$ f(x)=\frac{x^{4}}{x^{3}+1} $$

Step-by-Step Solution

Answer

The point at which the graph of function \(f(x)\) has a horizontal tangent is \((0,0)\)

Step 1: Compute the Derivative of \(f(x)\)

The original function is \(f(x)=\frac{x^{4}}{x^{3}+1}\). Applying the quotient rule (which states that the derivative of \(\frac{u}{v} = \frac{v(u') - u(v')}{v^2}\) where \(u = x^4\) and \(v = x^3 + 1\), the derivative \(f'(x)\) thus equals: \[f'(x)=\frac{(x^3+1)(4x^3)-(x^4)(3x^2)}{(x^3+1)^2}\] which simplifies as \[f'(x)= \frac{x^3}{((x^3+1)^2)}\].

Step 2: Set Derivative Equal to Zero and Solve for \(x\)

A horizontal tangent is where the slope (derivative) of the function equals zero. Here we have \[f'(x)= \frac{x^3}{((x^3+1)^2)}=0\] After cleaning up, we solve the equation \(x^3 = 0\), which gives us \(x = 0\)

Step 3: Substitute \(x\) into the Original Function \(f(x)\) to Find Corresponding \(y\) Values

Substitute \(x\) into the original function \(f(x)=\frac{x^{4}}{x^{3}+1}\) to find the corresponding \(y\). Thus, \(f(0) = \frac{0^{4}}{0^{3}+1} = 0\)

Step 4: Write the solution in \((x,y)\) format

The solution must be in the format \((x, y)\). Therefore, we have the point \((0,0)\)

Key Concepts

Derivative

In calculus, the derivative of a function measures the rate at which the function's value changes as its input changes. It's a fundamental concept that allows us to determine the slope or steepness of a curve at any given point. For a function like \( f(x) = \frac{x^4}{x^3 + 1} \), the derivative tells us how the function behaves as \( x \) changes, specifically how steep the curve is for each \( x \) value.

Finding the derivative involves calculus techniques, such as differentiation rules. The most basic rule is the power rule, applied to simple polynomials. However, when dealing with a quotient of functions, as in this exercise, we must apply the quotient rule. This leads us smoothly into our next topic.

Quotient Rule

The quotient rule is a technique for differentiating functions that are written as the ratio of two functions. If you have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable, the quotient rule states that the derivative \( y' \) is given by:

• \( y' = \frac{v(u') - u(v')}{v^2} \)

The roles of \( u \) and \( v \) make this formula both a method and a mnemonic: "minus the derivative of v times u!"

For \( f(x) = \frac{x^4}{x^3 + 1} \), we identify \( u = x^4 \) and \( v = x^3 + 1 \). By applying the quotient rule, we find the derivative is initially complicated, but simplifying gives us \( f'(x) = \frac{x^3}{(x^3+1)^2} \). This simplification is key to understanding how the graph behaves, particularly where the tangent is horizontal.

Algebraic Functions

Algebraic functions include polynomials, ratios of polynomials, and combinations of these through addition, subtraction, multiplication, division, and taking roots. They must not include any transcendental functions (e.g., trigonometric, exponential).

In this exercise, \( f(x) = \frac{x^4}{x^3 + 1} \) is an algebraic function because it is a ratio of two polynomial expressions. Understanding algebraic simplicity allows us to employ straightforward differentiation strategies, like the quotient rule. Recognizing this property guides us efficiently in analyzing the behavior of \( f(x) \) and aids in finding points like horizontal tangents.

Solving Equations

To determine where the function \( f(x) \) has a horizontal tangent, we set its derivative \( f'(x) \) equal to zero. A horizontal tangent reflects that the slope of the tangent line is zero. In terms of the derivative, this means:

• \( \frac{x^3}{(x^3+1)^2} = 0 \)

For such a fraction to equal zero, the numerator must equal zero since the denominator never becomes zero (as required for any valid function). Solving \( x^3 = 0 \), we find \( x = 0 \).

By substituting this back into the original function, \( f(x) \), we find the corresponding \( y \)-value and thus locate the exact point \((0,0)\) where the graph has a horizontal tangent. This method ensures a precise answer and deepens comprehension of where and why the function's slope changes to zero.