/*! 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 1 PROBLEMS. Let \(f(x)=1 /(1+x) ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 1

PROBLEMS. Let \(f(x)=1 /(1+x) .\) What is (i) \(\quad f(f(x))\) (for which \(x\) does this make sense?). (ii) \(\quad f\left(\frac{1}{x}\right)\) (iii) \(f(c x)\) (iv) \(\quad f(x+y)\) (v) \(\quad f(x)+f(y)\) (vi) For which numbers \(c\) is there a number \(x\) such that \(f(c x)=f(x)\) Hint: There are a lot more than you might think at first glance. (vii) For which numbers \(c\) is it true that \(f(c x)=f(x)\) for two different numbers \(x ?\)

Short Answer

Expert verified
(i) \(\frac{1+x}{2+x}\), valid for \(x eq -1\). (ii) \(\frac{x}{x+1}\). (iii) \(\frac{1}{1+cx}\). (iv) \(\frac{1}{1+x+y}\). (v) \(\frac{1}{1+x} + \frac{1}{1+y}\). (vi) \(c=1\). (vii) \(c \in \mathbb{R}\) for x=0, otherwise \(c=1\).

Step by step solution

01

Find f(f(x))

First, determine the expression for f(f(x)). Given that f(x) = \(\frac{1}{1+x}\), substitute f(x) into itself: \[ f(f(x)) = f\left(\frac{1}{1+x}\right) \] Now, apply f to the expression inside the parentheses:\[ f\left(\frac{1}{1+x}\right) = \frac{1}{1+\frac{1}{1+x}} \] Simplify the expression: \[ f\left(\frac{1}{1+x}\right) = \frac{1}{\frac{1+x+1}{x+1}} = \frac{1+x}{2+x} \] Therefore, \[ f(f(x)) = \frac{1+x}{2+x} \] This expression makes sense for any real number x, provided \(x eq -1\) so that the denominators are nonzero.
02

Find f(1/x)

Substitute \(\frac{1}{x}\) into the function f(x): \[ f\left(\frac{1}{x}\right) = \frac{1}{1 + \frac{1}{x}} \] Simplify the expression: \[ f\left(\frac{1}{x}\right) = \frac{1}{\frac{x+1}{x}} = \frac{x}{x+1} \] Therefore, \[ f\left(\frac{1}{x}\right) = \frac{x}{x+1} \]
03

Find f(c*x)

Substitute c*x into the function f(x): \[ f(c x) = \frac{1}{1 + c x} \] This expression is already simplified.
04

Find f(x+y)

Substitute x + y into the function f(x): \[ f(x + y) = \frac{1}{1 + x + y} \] This expression is already simplified.
05

Find f(x) + f(y)

Calculate f(x) and f(y) separately and then add them: \[ f(x) + f(y) = \frac{1}{1+x} + \frac{1}{1+y} \] This expression is already simplified.
06

Find numbers c such that f(c x) = f(x)

To find such c, equate f(c x) to f(x): \[ \frac{1}{1 + c x} = \frac{1}{1 + x} \] Cross-multiply to get: \[ 1 + x = 1 + c x \] Isolate c: \[ x = c x \] Divide both sides by x (assuming x eq 0): \[ c = 1 \] This means c must equal 1 to satisfy f(c x) = f(x) for all x.
07

Find numbers c for which f(c x) = f(x) for two different numbers x

Revisit the previous equation: \[ \frac{1}{1 + c x} = \frac{1}{1 + x} \] This must hold for two different values of x, say x1 and x2, both non-zero. Rewrite: \[ 1 + x1 = 1 + c x1 \] \[ 1 + x2 = 1 + c x2 \] Since the expressions must hold concurrently: \[ x1 = c x1 \] \[ x2 = c x2 \] This implies c can be 1 or any other value when x is 0: \[ c \in \mathbb{R} \] for x=0, any c satisfies. For non-zero x's, \(c = 1\).

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Key Concepts

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

Functional Composition
Functional composition occurs when you apply one function to the results of another function. For instance, given a function \(f(x) = \frac{1}{1+x}\), we can find \(f(f(x))\). Start by substituting \(f(x)\) back into itself: \(f(f(x)) = f\bigg(\frac{1}{1+x}\bigg)\). Next, apply \(f\) to the inside expression:
  • \(f\bigg(\frac{1}{1+x}\bigg) = \frac{1}{1+\frac{1}{1+x}}\)
  • Simplifies to \(\frac{1}{\frac{1+x+1}{1+x}} = \frac{1+x}{2+x}\).
Thus, \(f(f(x)) = \frac{1+x}{2+x}\), which makes sense for all real numbers \(x\) except \(x = -1\), where the denominator would be zero. This shows how we combine functions by nested substitutions to form a new one.
Inverse Functions
An inverse function essentially reverses the effect of the original function. To determine \(f(\frac{1}{x})\) for \(f(x) = \frac{1}{1+x}\), substitute \(\frac{1}{x}\) into \(f\) as follows: \(f\big(\frac{1}{x}\big) = \frac{1}{1+\frac{1}{x}}\), which simplifies to
  • \(\frac{1}{\frac{x+1}{x}} = \frac{x}{x+1}\).
Therefore, \(f\big(\frac{1}{x}\big) = \frac{x}{x+1}\). Understanding inverse functions is crucial as it allows us to find outputs that trace back to their original inputs through the inverse relationship.
Substitution in Functions
Substitution in functions involves replacing the variable in a function with another expression. For example, given \(f(x) = \frac{1}{1+x}\), we can find \(f(cx)\) by substituting \(cx\) for \(x\):
  • \(f(cx) = \frac{1}{1+cx}\).
This is a straightforward substitution that gives us an expression in terms of \(cx\). Similarly, finding \(f(x + y)\) involves substituting \(x + y\) into \(x\):
  • \(f(x + y) = \frac{1}{1 + x + y}\).
Having the skill to substitute expressions effectively is important for transforming and simplifying complex functions.
Simplification of Expressions
Simplification of expressions is a key step in solving many calculus problems. For instance, to simplify the sum \(f(x) + f(y)\) for \(f(x) = \frac{1}{1+x}\), you handle both functions separately and then add them up:
  • \(f(x) + f(y) = \frac{1}{1+x} + \frac{1}{1+y}\).
Moreover, finding \(c\) such that \(f(cx) = f(x)\) involves setting \(\frac{1}{1+cx} = \frac{1}{1+x}\) and solving for \(c\). This gives:
  • \(1 + x = 1 + cx\)
  • \(x = cx\)
  • \(c = 1\).
For \(c\) to work for two distinct values of \(x\), it generally means exploring more unique conditions. Generally, \(c = \text{1}\) is a critical insight, but exploring \(c eq 1\) and possible multi-value solutions is also beneficial. Simplification makes complex problems more manageable and solutions clearer.

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Most popular questions from this chapter

(a) Suppose \(g=h\) o \(f\). Prove that if \(f(x)=f(x)\), then \(g(x)=g(y)\) (b) Conversely suppose that \(f\) and \(g\) are two functions such that \(g(x)=g(y)\) whenever \(f(x)=f(y) .\) I'rove that \(g=h \circ f\) for some function \(h .\) Ilint: Just try to define \(h(z)\) when \(z\) is of the form \(z=f(x)\) (these are the only \(z\) that matter) and use the hypotheses to show that your definition will not run into trouble.

(a) Prove that there do not exist functions \(f\) and \(g\) with cither of the following properties: (i) \(\quad f(x)+g(y)=x y\) for all \(x\) and \(y\) (ii) \(f(x) \cdot g(y)=x+y\) for all \(x\) and \(y\) Hint: Try to get some information about \(f\) or \(g\) by choosing particular values of \(x\) and \(y\) (b) Find functions \(f\) and \(g\) such that \(f(x+y)=g(x y)\) for all \(x\) and \(y\)

(a) Prove that for any polynomial function \(f,\) and any number \(a,\) there is a polynomial function \(g,\) and a number \(b,\) such that \(f(x)=(x-a) g(x)+b\) for all \(x .\) (The idea is simply to divide \((x-a)\) into \(f(x)\) by long division. until a constant remainder is left. For example, the calculation (EQUATION CANNOT COPY) shows that \(x^{3}-3 x+1=(x-1)\left(x^{2}+x-2\right)-1 .\) A formal proof is possible by induction on the degree of \(f .\) ) (b) Prove that if \(f(a)=0 .\) then \(f(x)=(x-a) g(x)\) for some polynomial function \(g .\) (The converse is obvious.) (c) Prove that if \(f\) is a polynomial function of degree \(n\), then \(f\) has at most \(n\) roots, i.e., there are at most \(n\) numbers \(a\) with \(f(a)=0\). (d) Show that for each \(n\) there is a polynomial function of degree \(n\) with \(n\) roots. If \(n\) is even find a polynomial function of degree \(n\) with no roots, and if \(n\) is odd find one with only one root.

PROBLEMS. Let \(S(x)=x^{2},\) let \(P(x)=2^{x},\) and let \(s(x)=\sin x .\) Find each of the following. In each case you answer should be a mumber. (i) \((S \circ P)(y)\) (ii) \(\quad(S \circ s)(y)\) (iii) \(\quad(S \circ P \circ s)(t)+(s \circ P)(t)\) (iv) \(s\left(t^{3}\right)\)

PROBLEMS. Let \(g(x)=x^{2},\) and let $$ h(x)=\left\\{\begin{array}{ll} 0, & x \text { rational } \\ 1, & x \text { irrational. } \end{array}\right. $$ (i) For which \(y\) is \(h(y) \leq y ?\) (ii) For which \(y\) is \(h(y) \leq g(y) ?\) (iii) What is \(g(h(z))-h(z) ?\) (iv) For which \(w\) is \(g(w) \leq w ?\) (v) For which \(\varepsilon\) is \(g(g(\varepsilon))=g(\varepsilon) ?\)
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