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Chapter 1: Problem 82

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. \(f(x)=\frac{0.3 x}{0.04+x^{2}}\)

Short Answer

Expert verified
The tangent line to \( f(x) \) is horizontal at \( x = 0.2 \) and \( x = -0.2 \).

Step by step solution

01

Understand the concept of a horizontal tangent

For a tangent line to be horizontal, its slope must be zero. The slope of the tangent line to the function \( f(x) \) at any point \( x \) is given by the derivative \( f'(x) \). Therefore, we need to find the points where \( f'(x) = 0 \).
02

Differentiate the function

The function \( f(x) = \frac{0.3x}{0.04 + x^2} \) is a rational function. We use the quotient rule for differentiation: If \( u(x) = 0.3x \) and \( v(x) = 0.04 + x^2 \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).
03

Apply the quotient rule

- The derivative of the numerator \( u(x) = 0.3x \) is \( u'(x) = 0.3 \).- The derivative of the denominator \( v(x) = 0.04 + x^2 \) is \( v'(x) = 2x \).Therefore:\[ f'(x) = \frac{0.3(0.04 + x^2) - 0.3x(2x)}{(0.04 + x^2)^2} = \frac{0.012 + 0.3x^2 - 0.6x^2}{(0.04 + x^2)^2} \].This simplifies to:\[ f'(x) = \frac{0.012 - 0.3x^2}{(0.04 + x^2)^2} \].
04

Set the derivative to zero and solve for x

Set \( f'(x) = 0 \):\[ \frac{0.012 - 0.3x^2}{(0.04 + x^2)^2} = 0 \].This implies \( 0.012 - 0.3x^2 = 0 \).Solve for \( x \):\[ 0.3x^2 = 0.012 \]\[ x^2 = \frac{0.012}{0.3} = 0.04 \]\[ x = \pm \sqrt{0.04} \]\[ x = \pm 0.2 \].
05

Verify the solutions

The points \( x = 0.2 \) and \( x = -0.2 \) are valid as they arise from solving the equation \( 0.012 - 0.3x^2 = 0 \) and they lie within the domain of \( f(x) \). Thus, the tangent line to \( f(x) \) is horizontal at these points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
Understanding derivatives is fundamental to calculus. A derivative represents the rate at which a function is changing at any given point, and visually, it corresponds to the slope of the tangent line to the curve of the function at that point. For a function \( f(x) \), its derivative is denoted by \( f'(x) \). The process of finding this derivative is called differentiation.

In the context of the given problem, the derivative helps us find where the tangent line is horizontal by setting \( f'(x) = 0 \). This indicates points on the curve where there is no vertical change, meaning the slope is zero.
Tangent Line
A tangent line is a straight line that just "grazes" a curve at a specific point. It touches the curve without crossing it, much like a car tire touching the road. At the point of tangency, the tangent line and the curve have the same slope. This makes tangent lines incredibly useful for approximating the behavior of curves.

In the exercise, we are interested in finding points where the tangent line is horizontal. A horizontal tangent line has a slope of zero, which links directly to the derivative of the function being zero (since slope = derivative at point).
Quotient Rule
The quotient rule is a method for finding the derivative of a quotient of two differentiable functions. Given a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule states that:

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

This formula helps calculate the derivative of division-based functions easily. In the exercise, our function \( f(x) = \frac{0.3x}{0.04 + x^2} \) requires the quotient rule to find \( f'(x) \), as the function is a ratio of \( 0.3x \) and \( 0.04 + x^2 \). The derivative of each component, \( u \) and \( v \), is needed to apply this rule effectively.
Horizontal Tangent
A horizontal tangent line is a special case where the tangent line at a specific point on a curve is parallel to the x-axis. This means the slope of the line is zero. For polynomial and rational functions, like our given function \( f(x) = \frac{0.3x}{0.04 + x^2} \), we find horizontal tangents by setting the derivative to zero.

The equation obtained from setting \( f'(x) = 0 \) provides solutions for \( x \), which correspond to where the horizontal tangents occur. In the problem here, solving \( f'(x) = 0 \) led us to identify \( x = \pm 0.2 \) as points where the tangent line is horizontal, meaning there is no steepness or vertical tilt at these points.

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Most popular questions from this chapter

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 2}\left(\frac{x^{3}+5 x-18}{2 x^{2}-8}\right) $$

Find the interval(s) for which \(f^{\prime}(x)\) is positive. $$ f(x)=x^{2}-4 x+1 $$

Find the interval(s) for which \(f^{\prime}(x)\) is positive. Find the points on the graph of $$ y=x^{4}-\frac{4}{3} x^{2}-4 $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 10}\left(\frac{x^{2}+x-110}{x-10}\right) $$

Population growth rate. In \(t\) years, the population of Kingsville grows from 100,000 to a size \(P\) given by \(P(t)=100,000+2000 t^{2}\) a) Find the growth rate, \(d P / d t\). b) Find the population after 10 yr. c) Find the growth rate at \(t=10\). d) Explain the meaning of your answer to part (c).
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