/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 If \(f(x)\) is a function, then ... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(f(x)\) is a function, then \((f \circ f)(x)=f(f(x))\) is the composition of \(f\) with itself. This is called an iterated function, and the composition can be repeated many times. For example, \((f \circ f \circ f)(x)=f(f(f(x))) .\) Iterated functions are very useful in many areas, including finance (compound interest is a simple case) and the sciences (in weather forecasting, for example). For each function, use the Chain Rule to find the derivative $$ \text { If } f(x)=\sqrt{3 x}, \text { find } \frac{d}{d x}[(f \circ f \circ f)(x)] $$ Do you see a shortcut?

Short Answer

Expert verified
The derivative is \( \frac{3^{5/8}}{4}x^{-3/4} \). A shortcut is recognizing the pattern in iterated compositions form a power function, simplifying derivative calculations.

Step by step solution

01

Understanding the Iterated Function

Given the function \( f(x) = \sqrt{3x} \), we need to compute the iterated function \((f \circ f \circ f)(x)\). This means applying \( f \) three times: \( f(f(f(x))) \). Start by applying \( f \) once: \( f(x) = \sqrt{3x} \).
02

First Composition Step

Perform the first composition: \( f(f(x)) = f(\sqrt{3x}) = \sqrt{3(\sqrt{3x})}\). Simplify: \( f(f(x)) = \sqrt{3^{3/2}x^{1/2}} = (3^{1/4})\sqrt{x} \).
03

Second Composition Step

Perform the next composition: \( f(f(f(x))) = f((3^{1/4})\sqrt{x}) \). This results in \( \sqrt{3((3^{1/4})\sqrt{x})} = \sqrt{3^{5/4}x^{1/2}} = (3^{5/8})x^{1/4} \).
04

Applying the Chain Rule

To find the derivative \( \frac{d}{dx}[(f \circ f \circ f)(x)] \) where \( (f \circ f \circ f)(x) = (3^{5/8})x^{1/4} \): Use the derivative formula for power functions, \( \frac{d}{dx} x^n = nx^{n-1} \). For \( x^{1/4} \), the derivative is \( \frac{1}{4}x^{-3/4} \). Therefore, \( \frac{d}{dx}[(3^{5/8})x^{1/4}] = (3^{5/8}) \cdot \frac{1}{4}x^{-3/4} \).
05

Final Calculation of the Derivative

Evaluate the final expression for the derivative: \( \frac{3^{5/8}}{4} x^{-3/4} = \frac{3^{5/8}}{4\sqrt[4]{x^3}} \). Therefore, the final derivative is: \( \frac{3^{5/8}}{4}x^{-3/4} \).
06

Identifying a Shortcut

Notice that each time \( f \) is applied, the exponent and coefficient are modified consistently. By observing the pattern, one can derive similar compositions directly without recalculating the composition, recognizing that the function form (power function in a nested pattern) simplifies the derivative calculation.

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Key Concepts

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

Iterated Functions
Iterated functions involve applying a function repeatedly to its own output. In this concept, the function's result becomes the input for the next operation. This is akin to putting a result back into a machine to see what comes out next. For an iterated function, if we start with a function \( f(x) \), then \((f \circ f)(x)\) means \( f(f(x)) \), or applying \( f \) to \( x \) and then applying \( f \) again to that result. We can also have higher iterations like \((f \circ f \circ f)(x)\), which results in \( f(f(f(x))) \). Iterated functions appear in many real-world applications, from computing compound interest in finance to simulating weather patterns in meteorology. Each application of \( f \) often modifies the function in a predictable manner, providing a foundation for deeper mathematics and complex systems.
Composition of Functions
The composition of functions is a fundamental concept where two functions are combined in such a way that the output of one function becomes the input of another. If you have two functions, \( f(x) \) and \( g(x) \), the composition \((f \circ g)(x)\) means \( f(g(x)) \). We're essentially feeding the output of \( g(x) \) right into \( f(x) \). This process is similar to connecting two pipelines, where the output of the first is routed directly into the intake of the second. In mathematics, this operation is crucial for constructing new functions from existing ones, often leading to more complex and interesting results. Understanding this can be incredibly powerful not only in solving equations but also in modeling sophisticated real-life scenarios.
Chain Rule
The chain rule is vital in calculus for computing derivatives of composite functions. It allows us to find the derivative of a function that is composed of other functions. Imagine nesting one function inside another, much like a matryoshka doll, and wanting to find how sensitively the innermost doll (function) reacts to changes from the outside. For two functions, \( f(u) \) and \( u = g(x) \), the chain rule states that the derivative \( \frac{d}{dx}[f(g(x))] \) is \( f'(g(x)) \cdot g'(x) \). This means you find the derivative of \( f \) with respect to \( u \) and multiply it by the derivative of \( u \) with respect to \( x \). The chain rule is instrumental when handling iterated or composite functions, as it simplifies the differentiation process, providing a much clearer path to the solution.
Derivative Calculation
Calculating derivatives is a primary tool in calculus used to determine the rate of change of a function. When you have an expression like \((f \circ f \circ f)(x)\), accurate derivative calculation helps unravel how this complex function changes as \( x \) changes. For power functions, the derivative formula \( \frac{d}{dx} x^n = nx^{n-1} \) is particularly useful. In our exercise, where \( (f \circ f \circ f)(x) = (3^{5/8})x^{1/4} \), the derivative \( \frac{d}{dx}[(3^{5/8})x^{1/4}] = (3^{5/8}) \cdot \frac{1}{4}x^{-3/4} \) shows us the rate of change. Simplifying gives us \( \frac{3^{5/8}}{4\sqrt[4]{x^3}} \). Being able to calculate derivatives effectively allows us to analyze how function values shift concerning x, providing critical insights into their behavior.

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Most popular questions from this chapter

The equation $$ R(v)=\frac{6000}{v} $$ can be used to determine the heart rate, \(R,\) of a person whose heart pumps 6000 milliliters (mL) of blood per minute and v milliliters of blood per beat. a) Find the rate of change of heart rate with respect to \(v\), the output per beat. b) Find the heart rate at \(v=80 \mathrm{~mL}\) per beat. c) Find the rate of change at \(v=80 \mathrm{~mL}\) per beat.

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'Hôpital'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 first through the fourth derivatives. Be sure to simplify each derivative before continuing. $$ f(x)=\frac{x+3}{x-2} $$

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

It has been shown that the home range, in hectares, of a carnivorous mammal weighing \(w\) grams can be approximated by $$ H(w)=0.11 w^{1.36} $$ (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. \(216,\) Reading, MA: Addison-Wesley, 1973 ; and Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2, pp. 405-418.) a) Find the average rate at which a carnivorous mammal's home range increases as the animal's weight grows from \(500 \mathrm{~g}\) to \(700 \mathrm{~g}\). b) Find \(\frac{H(300)-H(200)}{300-200}\). What does this rate represent?

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