Chapter 2 — Functions and Dynamic Relations¶

2.12 — Model Exam¶

Duration: 100 minutes. Total score: 100 points. Allowed materials: pencil and paper; no calculator or formula sheet. Every result must be justified by showing the work: a correct answer without work does not receive full credit. When a function is given only by its formula, its domain is the implicit one and its codomain is \(\mathbb{R}\).

Part A — Direct computation (30 points, about 25 minutes)¶

Six independent items worth \(5\) points each.

A1. Determine the domain of \(f(x) = \dfrac{x+1}{\sqrt{4-x^2}}\).

A2. Let \(f(x)=x^2-1\) and \(g(x)=\dfrac{1}{x}\). Find \((f-g)(x)\) and its domain.

A3. Let \(g(x) = 2\sqrt{x+3}-4\). Identify the prototype it comes from and the parameters \(a\), \(h\), \(c\), \(k\) of the transformation, and determine \(\operatorname{Dom}(g)\) and \(\operatorname{Img}(g)\).

A4. Find the vertex of \(f(x)=3x^2-12x+7\) and classify it as an absolute maximum or minimum.

A5. Determine whether \(f(x)=x^3+x\) is even, odd, or neither.

A6. Let \(f(x)=x-2\) and \(g(x)=x^2\). Compute \((g\circ f)(x)\) and \((f\circ g)(x)\), and verify that they are different functions.

Part B — Complete analysis of a function (25 points, about 25 minutes)¶

Let \(f:[2,\infty)\to\mathbb{R}\) be given by \(f(x) = (x-2)^2+3\).

(i) [4 pts] State \(\operatorname{Dom}(f)\) and find \(\operatorname{Img}(f)\).

(ii) [4 pts] Determine whether it makes sense to ask if \(f\) is even or odd on this domain, and justify.

(iii) [5 pts] Prove, with the definition of monotonicity and not with the graph, that \(f\) is increasing on \([2,\infty)\).

(iv) [4 pts] Determine whether \(f\) is injective, justifying with the appropriate test.

(v) [8 pts] Regarding \(f\) as a function from \([2,\infty)\) to its image, find \(f^{-1}\), state its domain and image, and verify the identity \(f^{-1}(f(x))=x\) for \(x=5\).

Part C — Composition and its domain (15 points, about 15 minutes)¶

Let \(f(x) = x^2-4\) and \(g(x)=\sqrt{x}\).

(i) [5 pts] Find \((g\circ f)(x)\) and its domain.

(ii) [5 pts] Find \((f\circ g)(x)\) and its domain.

(iii) [5 pts] The simplified formula for \((f\circ g)(x)\) obtained in (ii) seems to make sense for every \(x\in\mathbb{R}\). Explain precisely why the correct domain is not \(\mathbb{R}\).

Part D — Proof (15 points, about 20 minutes)¶

D1. [15 pts] Prove that if \(f\) is an odd function with \(0\in\operatorname{Dom}(f)\), then \(f(0)=0\).

Part E — Applied problem (15 points, about 15 minutes)¶

An object is thrown vertically upward. Its height, in meters, \(t\) seconds after the throw, is given by

\[h(t) = -5t^2+20t+1, \qquad t\ge0.\]

(i) [3 pts] Identify \(a\), \(b\) and \(c\), and determine whether \(h\) has an absolute maximum or minimum.

(ii) [7 pts] Compute the instant at which the object reaches its extreme height and the value of that height.

(iii) [5 pts] Check the result in the context of the problem: justify that the instant found belongs to the domain \(t\ge0\) and that the height obtained is physically reasonable.

Grading guide¶

Part A¶

A1. 2 pts for setting up \(4-x^2>0\), with strict inequality because the root is in the denominator; 2 pts for solving it; 1 pt for interval notation. Answer: \((-2,2)\).
A2. 2 pts for \((f-g)(x)=x^2-1-\dfrac{1}{x}\); 3 pts for the domain \(\mathbb{R}\cap(\mathbb{R}\setminus\{0\})=\mathbb{R}\setminus\{0\}\), justified as an intersection.
A3. 2 pts for the prototype \(\sqrt{x}\) and the four parameters \(a=1\), \(h=-3\), \(c=2\), \(k=-4\); a wrong prototype invalidates these 2 pts. 1 pt for \(\operatorname{Dom}(g)=[-3,\infty)\); 2 pts for \(\operatorname{Img}(g)=[-4,\infty)\), justified from \(\operatorname{Img}(\sqrt{x})=[0,\infty)\).
A4. 2 pts for \(x=-\frac{-12}{2\cdot3}=2\); 2 pts for \(f(2)=-5\); 1 pt for the classification: absolute minimum, since \(3>0\).
A5. 3 pts for computing \(f(-x)=-x^3-x\); 2 pts for the conclusion, odd, with the explicit comparison with \(-f(x)=-x^3-x\) and the remark that \(\operatorname{Dom}(f)=\mathbb{R}\) is symmetric.
A6. 2 pts for \((g\circ f)(x)=(x-2)^2\); 2 pts for \((f\circ g)(x)=x^2-2\); 1 pt for showing that they differ at a specific value, for instance at \(x=0\), where they equal \(4\) and \(-2\). Computing both without that conclusion does not earn this point.

Part B¶

(i) 1 pt for the domain \([2,\infty)\); 3 pts for \(\operatorname{Img}(f)=[3,\infty)\), proving both inclusions: \(f(x)\ge3\) because \((x-2)^2\ge0\), and each \(y\ge3\) is attained at \(x=2+\sqrt{y-3}\).
(ii) 4 pts only if the student recognizes that \([2,\infty)\) is not symmetric about the origin, so parity, defined in §2.3 for symmetric domains, does not apply. Computing \(f(-x)\) without pointing out this prior problem earns at most \(1\) pt.
(iii) 2 pts for taking \(2\le x_1<x_2\) and comparing \(f(x_1)\) with \(f(x_2)\); 3 pts for concluding \(f(x_1)<f(x_2)\) without appealing to the graph, for example: \(0\le x_1-2<x_2-2\) implies \((x_1-2)^2<(x_2-2)^2\), by the order lemma of §1.2, and \(3\) is added.
(iv) 2 pts for the correct answer, yes it is injective; 2 pts for justifying it with the monotonicity proved in (iii) or with the horizontal-line test.
(v) 4 pts for solving up to \(f^{-1}(x)=2+\sqrt{x-3}\), choosing the positive root because \(x\ge2\); 2 pts for the domain \([3,\infty)\) and the image \([2,\infty)\); 2 pts for the check: \(f(5)=12\) and \(f^{-1}(12)=2+\sqrt{9}=5\).

Part C¶

(i) 3 pts for \((g\circ f)(x)=\sqrt{x^2-4}\); 2 pts for the domain \((-\infty,-2]\cup[2,\infty)\).
(ii) 3 pts for \((f\circ g)(x)=(\sqrt{x})^2-4=x-4\); 2 pts for the domain \([0,\infty)\). Answering \(\mathbb{R}\) earns \(0\) pts on this part.
(iii) The \(5\) pts are awarded only for the argument: the domain of \(f\circ g\) requires \(x\in\operatorname{Dom}(g)\), that is \(x\ge0\), before applying \(f\), and the later algebraic simplification does not recover that restriction; it is the same phenomenon as the simplifications of §1.4 and the equality of functions of §2.2. An answer that only says "because that is what the computation gives" earns \(0\) pts.

Part D¶

5 pts for applying the definition of odd function at \(x=0\), which is valid because \(0\in\operatorname{Dom}(f)\): \(f(-0)=-f(0)\).
5 pts for explicitly recognizing that \(-0=0\) and therefore \(f(-0)=f(0)\).
5 pts for combining both equalities into \(f(0)=-f(0)\), concluding \(2f(0)=0\) and obtaining \(f(0)=0\).

Part E¶

(i) 2 pts for \(a=-5\), \(b=20\), \(c=1\); 1 pt for the classification, absolute maximum, only if justified by the sign of \(a\).
(ii) 3 pts for \(t=-\dfrac{b}{2a}=-\dfrac{20}{2\cdot(-5)}=2\); 4 pts for \(h(2)=-20+40+1=21\), with the explicit computation.
(iii) 2 pts for checking that \(t=2\ge0\), so that the vertex lies in the domain and the maximum of the formula is also the maximum of the problem; 3 pts for arguing that \(21\) m is a reasonable height, positive and greater than the initial height \(h(0)=1\) m.

With the model exam and its grading guide, Chapter 2 completes its cycle: theoretical development, synthesis, self-assessment by levels and a final assessment instrument. The following chapters take up this same apparatus, domain, transformation, composition and inverse, and apply it to specific families of functions.