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Chapter 2 — Functions and Dynamic Relations

2.11 — Review Problems

The problems below do not repeat the examples solved in the chapter: each one requires, at some point, a decision the reader must make alone, whether which technique to apply, in what order, or why a tool is not enough. They are organized in the three levels of the competency checklist of §2.10. The reference in brackets indicates the section where the necessary tools are found; several problems require combining more than one.

We use the chapter's notation: \(\operatorname{Dom}(f)\) is the domain of \(f\), \(\operatorname{Img}(f)\) its image and \(\operatorname{Gr}(f)\) its graph. When a function is given only by its formula, its domain is the implicit one and its codomain is \(\mathbb{R}\), following the convention of §2.2.


Level 1 — Mechanical and technical

1. [§2.2] Determine the implicit domain of \(f(x) = \dfrac{\sqrt{x+5}}{x^2-9}\).

2. [§2.1, §2.3] Let \(f(x)=3x-1\), with explicit domain \([0,4]\). Using that \(f\) is increasing, find \(\operatorname{Img}(f)\).

3. [§2.3] Determine whether \(f(x)=x^4-3x^2\) is even, odd, or neither.

4. [§2.4] Without recomputing from scratch, state the domain, image and monotonicity of \(f(x)=|x|\), relying on the catalog.

5. [§2.5] Describe in words the transformation that takes \(y=x^2\) to \(y=-3(x+2)^2-1\), identifying \(a\), \(h\), \(c\) and \(k\).

6. [§2.6] Given \(f(x)=x+2\) and \(g(x)=\sqrt{x}\), find \((f+g)(x)\), \((f\cdot g)(x)\) and the domain of each.

7. [§2.6] With the same \(f\) and \(g\) as in the previous problem, find \((g\circ f)(x)\) and its domain.

8. [§2.7] Determine whether \(f(x)=x^2+1\), with domain \([0,\infty)\), is injective, and if so find \(f^{-1}\).

9. [§2.8] Find the vertex of \(f(x)=-2x^2+8x-3\) and classify it as an absolute maximum or minimum.


Level 2 — Integration and proof

1. [§2.2, §2.6] Let \(h(x) = \dfrac{\sqrt{x-2}}{x-5}\). Compute \(\operatorname{Dom}(h)\) directly, and check the result by writing \(h\) as a quotient of two known functions and applying the rule of §2.6 for the domain of a quotient.

2. [§2.3] Prove, with the definition of monotonicity and without appealing to the graph, that \(f(x)=\dfrac{1}{x}\) is decreasing on \((0,\infty)\).

3. [§2.3, §2.5] Let \(f(x)=x^2-4\) and let \(g(x)=f(-x)\). Check, by computing \(g(x)\), that \(g = f\), and therefore \(\operatorname{Gr}(g)=\operatorname{Gr}(f)\). Explain which result of §2.5 guarantees this in advance.

4. [§2.6] Let \(f(x)=2x\) and \(g(x)=\sqrt{x-3}\). Compute \(\operatorname{Dom}(g\circ f)\) and explain, with the definition of composition of §2.6, why considering \(\operatorname{Dom}(f)\) alone is not enough.

5. [§2.7] Prove that \(f(x) = \dfrac{2x+1}{x-3}\) is injective on its domain, find its image, find \(f^{-1}\) and verify the identity \(f^{-1}(f(x))=x\).

6. [§2.8] Among all pairs of positive reals whose product is \(36\), find the pair that minimizes the sum, using the AM-GM inequality. Check the result by evaluating the sum for at least two different pairs that satisfy the constraint.

7. [§2.1, §2.3] Let \(R=\{(x,y)\in\mathbb{R}\times\mathbb{R} \mid y^2=x\}\). Determine whether \(R\) is a function from \([0,\infty)\) to \(\mathbb{R}\). If it is not, propose a subset of \(R\) that is a function from \([0,\infty)\) to \(\mathbb{R}\), and find its image.


Level 3 — Challenge, application and frontier

1. [§2.4] Formalize and prove, with the same \(M\) and \(\delta\) scheme of §2.4, that \(f(x)=\dfrac{2}{x-3}\) grows without bound as \(x\to3^+\).

2. [§2.8] The function \(f(x)=x^3-3x\) is, according to the frontier note of §2.8, a case that none of the tools of that section fully settles. By evaluating \(f\) on a grid of points of \([-2,2]\), determine as precisely as possible on which subintervals it appears to increase and on which it appears to decrease. Discuss why this method, though informative, cannot guarantee the exact location of any relative extremum.

3. [§2.6, §2.7] Let \(f:A\to B\) and \(g:B\to C\) be bijective functions. Prove that \(g\circ f:A\to C\) is bijective and that \((g\circ f)^{-1} = f^{-1}\circ g^{-1}\).

4. [§1.6, §2.8] A farmer has \(200\) meters of wire to fence a rectangular pen, using a straight wall as one of the four sides, so that the wire covers only the other three. Set up the area as a function of a single side, following the problem-solving method of §1.6; find the dimensions that maximize the area with the tools of §2.8, and check the result in the context of the problem.

5. [§2.1, §2.2] An equivalence relation on a set \(X\) is a relation \(\sim\) between elements of \(X\) that is reflexive (\(u\sim u\) for every \(u\)), symmetric (if \(u\sim v\), then \(v\sim u\)) and transitive (if \(u\sim v\) and \(v\sim w\), then \(u\sim w\)). Prove that the equality of functions defined in §2.2, according to which \(f:A\to B\) and \(g:C\to D\) are equal when \(A=C\), \(B=D\) and \(f(x)=g(x)\) for every \(x\in A\), is an equivalence relation on any set of functions.


Answer key

The answers for levels 1 and 2 are given in full when the result is a value or a closed expression; when the problem asks for a proof, the central step is indicated. The answers for level 3 are hints rather than solutions, so as not to empty each problem of its challenge.

Level 1

  1. We need \(x+5\ge0\) and \(x^2-9\neq0\), that is, \(x\ge-5\) and \(x\neq\pm3\): \(\operatorname{Dom}(f) = [-5,-3)\cup(-3,3)\cup(3,\infty)\).
  2. Since \(f\) is increasing, \(f(0)=-1\le f(x)\le f(4)=11\) for \(x\in[0,4]\); and every \(y\in[-1,11]\) is attained at \(x=\frac{y+1}{3}\in[0,4]\). Hence \(\operatorname{Img}(f) = [-1,11]\).
  3. Even: \(f(-x)=(-x)^4-3(-x)^2=x^4-3x^2=f(x)\) for every \(x\in\mathbb{R}\).
  4. \(\operatorname{Dom}=\mathbb{R}\), \(\operatorname{Img}=[0,\infty)\); decreasing on \((-\infty,0]\) and increasing on \([0,\infty)\).
  5. \(a=1\), \(h=-2\), \(c=-3\), \(k=-1\): vertical stretch by a factor of \(3\) with reflection across the \(x\)-axis, shift \(2\) units to the left and \(1\) unit down.
  6. \((f+g)(x) = x+2+\sqrt{x}\) and \((f\cdot g)(x) = (x+2)\sqrt{x}\), both with domain \(\mathbb{R}\cap[0,\infty)=[0,\infty)\).
  7. \((g\circ f)(x) = \sqrt{x+2}\), with domain \(\{x \mid x+2\ge0\}=[-2,\infty)\).
  8. Injective, since it is increasing on \([0,\infty)\); its image is \([1,\infty)\) and \(f^{-1}(x) = \sqrt{x-1}\), with domain \([1,\infty)\).
  9. Vertex at \(x=-\frac{8}{2\cdot(-2)}=2\), with \(f(2)=5\); absolute maximum, since \(-2<0\).

Level 2

  1. \(\operatorname{Dom}(h) = [2,5)\cup(5,\infty)\). With \(u(x)=\sqrt{x-2}\) and \(v(x)=x-5\), the rule gives \(\operatorname{Dom}(u)\cap\operatorname{Dom}(v)\) minus the zeros of \(v\): \([2,\infty)\setminus\{5\}\).
  2. For \(0<x_1<x_2\): \(\dfrac{1}{x_1}-\dfrac{1}{x_2} = \dfrac{x_2-x_1}{x_1x_2}>0\), because numerator and denominator are positive; hence \(f(x_1)>f(x_2)\).
  3. \(g(x)=(-x)^2-4=x^2-4=f(x)\), with the same domain \(\mathbb{R}\). Section 2.5 shows that \(\operatorname{Gr}(g)\) is the reflection of \(\operatorname{Gr}(f)\) across the \(y\)-axis, and an even function coincides with its reflection.
  4. \((g\circ f)(x)=\sqrt{2x-3}\), and \(\operatorname{Dom}(g\circ f) = \{x\in\mathbb{R} \mid 2x\ge3\}=\left[\frac{3}{2},\infty\right)\), strictly contained in \(\operatorname{Dom}(f)=\mathbb{R}\): the definition also requires \(f(x)\in\operatorname{Dom}(g)\).
  5. If \(f(x_1)=f(x_2)\), cross-multiplying gives \((2x_1+1)(x_2-3)=(2x_2+1)(x_1-3)\), which reduces to \(7x_1=7x_2\). Solving \(y=\frac{2x+1}{x-3}\) gives \(x=\frac{3y+1}{y-2}\), possible if and only if \(y\neq2\); hence \(\operatorname{Img}(f)=\mathbb{R}\setminus\{2\}\) and, as a function from \(\mathbb{R}\setminus\{3\}\) to \(\mathbb{R}\setminus\{2\}\), \(f^{-1}(x) = \dfrac{3x+1}{x-2}\).
  6. By AM-GM, \(a+b\ge2\sqrt{ab}=12\), with equality if and only if \(a=b=6\). For example, \(4+9=13>12\) and \(3+12=15>12\).
  7. No: \((4,2)\) and \((4,-2)\) are in \(R\), and uniqueness fails. The subset \(\{(x,\sqrt{x}) \mid x\ge0\}\) is a function, with image \([0,\infty)\); so is \(\{(x,-\sqrt{x}) \mid x\ge0\}\), with image \((-\infty,0]\).

Level 3 — hints

  1. Given \(M>0\), choose \(\delta=\frac{2}{M}\): if \(0<x-3<\delta\), then \(\frac{2}{x-3}>\frac{2}{\delta}=M\).
  2. At \(x=-2,\ -1.5,\ -1,\ -0.5,\ 0,\ 0.5,\ 1,\ 1.5,\ 2\) we get \(-2,\ 1.125,\ 2,\ 1.375,\ 0,\ -1.375,\ -2,\ -1.125,\ 2\). The values suggest growth up to near \(x=-1\), decrease up to near \(x=1\) and growth afterwards. A finite number of evaluations says nothing about what happens between two grid points: it only bounds where the extremum can be.
  3. Injectivity: if \(g(f(x_1))=g(f(x_2))\), the injectivity of \(g\) gives \(f(x_1)=f(x_2)\), and that of \(f\) gives \(x_1=x_2\). Surjectivity: given \(z\in C\), there is \(y\in B\) with \(g(y)=z\) and there is \(x\in A\) with \(f(x)=y\). For the inverse, use the associativity of §2.6 and check that \((f^{-1}\circ g^{-1})\circ(g\circ f)=\mathrm{id}_A\) and \((g\circ f)\circ(f^{-1}\circ g^{-1})=\mathrm{id}_C\); the characterization of §2.7 concludes.
  4. With \(x\) the side perpendicular to the wall, the parallel side is \(200-2x\) and the area is \(A(x)=x(200-2x)=-2x^2+200x\), with \(0<x<100\). The vertex is at \(x=50\): the parallel side measures \(100\) m and the maximum area is \(5000\) m². Both sides are positive and exactly \(50+50+100=200\) m of wire are used.
  5. Reflexivity and symmetry follow directly from the properties of equality of sets and of numbers. For transitivity, if \(f=g\) and \(g=k\), chain the equalities of domains, those of codomains and, for each \(x\) of the common domain, \(f(x)=g(x)=k(x)\).


With these problems solved, or at least seriously attempted, including those of level 3 that do not always yield on the first try, the reader is ready to take the chapter exam, which assesses, under time pressure, a sample of these same three levels.