Problem 81 - 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. \(f(x)=\frac{0.01 x^{2}}{x^{4}+0.0256}\)

Step-by-Step Solution

Answer

Horizontal tangents at \(x = 0, 0.4, -0.4\).

Step 1: Understand Horizontal Tangents

A horizontal tangent occurs where the derivative of the function is equal to zero. So, we need to find the derivative of the given function and set it equal to zero.

Step 2: Find the Derivative

The function given is \(f(x)=\frac{0.01 x^{2}}{x^{4}+0.0256}\). Use the quotient rule to differentiate: \[ f'(x) = \frac{(x^4+0.0256)(2\cdot0.01x) - 0.01x^2(4x^3)}{(x^4+0.0256)^2} \]. Simplify this expression to find a more workable form.

Step 3: Simplify Derivative

Simplify the derivative expression: \[ f'(x) = \frac{0.02x(x^4+0.0256) - 0.04x^5}{(x^4+0.0256)^2} = \frac{0.02x^5 + 0.02x\cdot0.0256 - 0.04x^5}{(x^4+0.0256)^2} \]. Further simplify to \[ f'(x) = \frac{-0.02x^5 + 0.000512x}{(x^4+0.0256)^2} \].

Step 4: Set Derivative to Zero

For a horizontal tangent, set the numerator of the derivative equal to zero: \(-0.02x^5 + 0.000512x = 0\). Factor out \(x\): \(x(-0.02x^4 + 0.000512) = 0\). This implies that \(x = 0\) or \(-0.02x^4 + 0.000512 = 0\).

Step 5: Solve for x

Solving \(-0.02x^4 + 0.000512 = 0\): \(-0.02x^4 = -0.000512\), divide both sides by \(-0.02\) to get \(x^4 = 0.0256\). Taking the 4th root of both sides gives \(x = \pm 0.4\).

Step 6: Conclusion

The points at which the tangent to \(f(x)\) is horizontal are \(x = 0\), \(x = 0.4\), and \(x = -0.4\). Ensure these values are not causing undefined behavior in the original function.

Key Concepts

Quotient Rule

The quotient rule is a method used in calculus to find the derivative of a function that is divided by another function. In this context, we have a function in the form of a quotient:

• Numerator: \( 0.01x^2 \)
• Denominator: \( x^4 + 0.0256 \)

When you take the derivative of a quotient, you can't simply take the derivative of the top and bottom separately. Instead, you use the quotient rule, which is: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]where \( u \) is the numerator, \( v \) is the denominator, and \( u' \) and \( v' \) are their respective derivatives. Applying the quotient rule makes it easier to handle the complexity of a function divided by another.

Derivative Simplification

After applying the quotient rule, we often end up with a complex expression. Derivative simplification is the process of reducing this expression to its simplest form. This makes subsequent calculations more manageable.
In the solution, the derivative of the function was: \[ f'(x) = \frac{0.02x(x^4+0.0256) - 0.04x^5}{(x^4+0.0256)^2} \]Simplifying involved combining like terms:

• Combine \( 0.02x^5 \) and \(-0.04x^5\)
• Factor common factors where possible

The simplified derivative becomes: \[ f'(x) = \frac{-0.02x^5 + 0.000512x}{(x^4+0.0256)^2}.\]Simplifying your derivative not only makes it easier to work with but also more straightforward to set up equations, like solving for horizontal tangents.

Fourth Root

Taking roots is an essential part of solving equations in calculus, especially when dealing with powers. In our exercise, to find the critical points, we needed to solve an equation that resulted in an expression like this: \( x^4 = 0.0256 \).
To isolate \( x \), we take the fourth root of both sides. The fourth root is essentially asking, "What number multiplied by itself four times gives me this value?"

• Fourth root of \( 0.0256 \) is \( \pm 0.4 \)

The "\( \pm \)" indicates that both positive and negative numbers can be roots here. Taking the fourth root is crucial for solving higher-degree polynomials that appear in derivative equations when determining possible points of horizontal tangency.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are important because they can indicate local maxima, minima, or points of inflection. In the exercise, we found the critical points by solving:
\[ -0.02x^5 + 0.000512x = 0 \] We factor to simplify:

• Take \( x \) common: \( x(-0.02x^4 + 0.000512) = 0 \)
• Solutions: \( x = 0 \) or solve \( -0.02x^4 + 0.000512 = 0 \)

For \( x = 0 \), it provides an immediate critical point. For the second part, solving \( x^4 = 0.0256 \) gives the critical points: \( x = \pm 0.4 \).
Critical points are crucial as they guide us to locations where the function behavior might change, like where a tangent horizontal to the curve might exist.