Functions and Their Graphs — Worked Solutions

Ques­tion 1: Which rela­tions are func­tions?#

The prob­lem#

Which of these rela­tions on $\{1, 2, 3, 4\}$ are func­tions?

(a) $\{(1, 3), (2, 3), (3, 3), (4, 3)\}$;

(b) $\{(1, 1), (2, 1), (2, 4), (3, 2), (4, 3)\}$;

(c) $\{(1, 4), (2, 3), (3, 2)\}$.

Under­stand­ing the prob­lem#

Each rela­tion is a set of ordered pairs whose first ele­ments should come from the domain $\{1, 2, 3, 4\}$. You must decide, for each, whether it is a func­tion on this set.

The idea#

A rela­tion is a func­tion on $A$ when every ele­ment of $A$ appears as a first ele­ment exactly once. So check two things: (i) no ele­ment is miss­ing; (ii) no ele­ment has two dif­fer­ent images. It is fine for dif­fer­ent ele­ments to share the same image.

Step-by-step solu­tion#

Part (a)

Step 1. List the first ele­ments: $1, 2, 3, 4$. Each appears once, and none is miss­ing.

Step 2. All four share the image $3$. That is allowed (like Exam­ple 1 in the les­son). So (a) is a func­tion — in fact a con­stant func­tion.

Part (b)

Step 1. List the first ele­ments: $1, 2, 2, 3, 4$. The ele­ment $2$ appears twice, with images $1$ and $4$.

Step 2. An ele­ment with two images breaks the rule, so (b) is not a func­tion.

Part (c)

Step 1. List the first ele­ments: $1, 2, 3$. The ele­ment $4$ never appears.

Step 2. Every ele­ment of the domain must have an image, so (c) is not a func­tion on $\{1, 2, 3, 4\}$.

Check­ing the answer#

Think of an arrow dia­gram from $\{1,2,3,4\}$: in (a) exactly one arrow leaves each ele­ment; in (b) two arrows leave $2$; in (c) no arrow leaves $4$.

Answer#

(a) Yes; (b) no, because $2$ has two images; (c) no, because $4$ has no image.

Ques­tion 2: Eval­u­at­ing a func­tion#

The prob­lem#

Let $f(x) = x^2 - 4x$. Find $f(0)$, $f(-1)$, $f(4)$ and $f(a + 1)$.

Under­stand­ing the prob­lem#

To eval­u­ate $f$ at any input, replace every $x$ in the rule by that input. The last one has a let­ter, $a + 1$, as the input — the answer will be an expres­sion in $a$.

The idea#

Direct sub­sti­tu­tion, using brack­ets around the input so that signs and squares are han­dled cor­rectly.

Step-by-step solu­tion#

Step 1. $x = 0$:

$$f(0) = 0^2 - 4(0) = 0.$$

Step 2. $x = -1$ (the bracket mat­ters: $(-1)^2 = 1$):

$$f(-1) = (-1)^2 - 4(-1) = 1 + 4 = 5.$$

Step 3. $x = 4$:

$$f(4) = 16 - 16 = 0.$$

Step 4. $x = a + 1$:

$$\begin{aligned} f(a + 1) &= (a + 1)^2 - 4(a + 1) \\ &= a^2 + 2a + 1 - 4a - 4 \\ &= a^2 - 2a - 3. \end{aligned}$$

Check­ing the answer#

Put $a = -1$ in $a^2 - 2a - 3$: $1 + 2 - 3 = 0$, which should equal $f(0) = 0$. It does. Put $a = 3$: $9 - 6 - 3 = 0 = f(4)$. It does.

Answer#

$f(0) = 0$, $f(-1) = 5$, $f(4) = 0$, $f(a + 1) = a^2 - 2a - 3$.

Com­mon mis­take to avoid#

Writ­ing $-1^2 = -1$ instead of $(-1)^2 = 1$. Always bracket a neg­a­tive input.

Ques­tion 3: Great­est inte­ger and mod­u­lus val­ues#

The prob­lem#

Eval­u­ate: $[3.9]$; $[-0.4]$; $[-6]$; $\lvert -7 \rvert - \lvert 2 \rvert$.

Under­stand­ing the prob­lem#

$[x]$ is the great­est inte­ger not exceed­ing $x$ — the inte­ger at or imme­di­ately to the left of $x$ on the num­ber line. $\lvert x \rvert$ is the dis­tance of $x$ from $0$.

The idea#

Pic­ture each num­ber on the num­ber line and move left to the near­est inte­ger (or stay, if it already is an inte­ger).

Step-by-step solu­tion#

Step 1. $3.9$ lies between $3$ and $4$; the great­est inte­ger not exceed­ing it is $3$. So $[3.9] = 3$.

Step 2. $-0.4$ lies between $-1$ and $0$. The inte­ger to its left is $-1$ ($0$ is big­ger than $-0.4$, so it is not allowed). So $[-0.4] = -1$.

Step 3. $-6$ is already an inte­ger, so $[-6] = -6$.

Step 4. $\lvert -7 \rvert = 7$ and $\lvert 2 \rvert = 2$, so

$$\lvert -7 \rvert - \lvert 2 \rvert = 7 - 2 = 5.$$

Check­ing the answer#

In each case $[x] \le x < [x] + 1$: $3 \le 3.9 < 4$; $-1 \le -0.4 < 0$; $-6 \le -6 < -5$. All hold.

Answer#

$3$; $-1$; $-6$; $5$.

Com­mon mis­take to avoid#

Answer­ing $[-0.4] = 0$ by "drop­ping the dec­i­mal". For neg­a­tive num­bers the great­est inte­ger is the next inte­ger down.

Ques­tion 4: Find­ing domains#

The prob­lem#

Find the domain of

(a) $\displaystyle \frac{1}{x^2 - 9}$; (b) $\sqrt{8 - 2x}$; (c) $\displaystyle \frac{x}{x^2 + 4}$.

Under­stand­ing the prob­lem#

The domain of a real func­tion given by a for­mula is the set of all real $x$ for which the for­mula gives a real num­ber.

The idea#

Two restric­tions mat­ter here: a denom­i­na­tor can­not be zero, and the expres­sion under a square root can­not be neg­a­tive.

Step-by-step solu­tion#

Part (a)

Step 1. Find where the denom­i­na­tor is zero.

$$x^2 - 9 = 0 \;\Longrightarrow\; (x - 3)(x + 3) = 0 \;\Longrightarrow\; x = 3 \text{ or } x = -3.$$

Step 2. Every other real num­ber works, so the domain is $\mathbf{R} - \{-3, 3\}$.

Part (b)

Step 1. Require the expres­sion under the root to be non-neg­a­tive.

$$8 - 2x \ge 0 \;\Longrightarrow\; 8 \ge 2x \;\Longrightarrow\; x \le 4.$$

Step 2. The domain is $(-\infty, 4]$. (At $x = 4$ the value is $\sqrt{0} = 0$, which is allowed.)

Part (c)

Step 1. Check whether the denom­i­na­tor can be zero: $x^2 + 4 = 0$ would need $x^2 = -4$, impos­si­ble for real $x$, because $x^2 \ge 0$ so $x^2 + 4 \ge 4$.

Step 2. No value is excluded; the domain is $\mathbf{R}$.

Check­ing the answer#

Test $x = 5$ in (b): $8 - 10 = -2 < 0$, no real root — out­side the domain, as expected. Test $x = 0$: $\sqrt 8$ is real — inside.

Answer#

(a) $\mathbf{R} - \{-3, 3\}$; (b) $(-\infty, 4]$; (c) $\mathbf{R}$.

Ques­tion 5: Find­ing ranges#

The prob­lem#

Find the range of

(a) $x^2 + 5$; (b) $\lvert x + 1 \rvert - 3$; (c) $\displaystyle \frac{1}{x^2 + 1}$.

Under­stand­ing the prob­lem#

The range is the set of all out­put val­ues the func­tion actu­ally takes as $x$ runs over its domain (here all of $\mathbf{R}$ in each case).

The idea#

Start from a basic fact — $x^2 \ge 0$ and $\lvert t \rvert \ge 0$, both tak­ing every non-neg­a­tive value — and build the func­tion step by step, track­ing what hap­pens to the set of val­ues.

Step-by-step solu­tion#

Part (a)

Step 1. $x^2$ takes every value in $[0, \infty)$.

Step 2. Adding $5$ shifts every value up by $5$: $x^2 + 5$ takes every value in $[5, \infty)$. The small­est value $5$ occurs at $x = 0$.

Part (b)

Step 1. $\lvert x + 1 \rvert$ takes every value in $[0, \infty)$ (it is $0$ at $x = -1$ and grows with­out bound).

Step 2. Sub­tract­ing $3$ shifts down by $3$: the range is $[-3, \infty)$.

Part (c)

Step 1. $x^2 + 1$ takes every value in $[1, \infty)$.

Step 2. Tak­ing the rec­i­p­ro­cal of a num­ber $t \ge 1$ gives $\displaystyle 0 < \tfrac1t \le 1$. When $t = 1$ (at $x = 0$) we get $1$; as $t$ grows, $\displaystyle \tfrac1t$ gets close to $0$ but never reaches it.

Step 3. So the range is $(0, 1]$.

Check­ing the answer#

For (c), $x = 0$ gives $1$, $x = 1$ gives $\displaystyle \tfrac12$, $x = 10$ gives $\displaystyle \tfrac{1}{101}$: all in $(0, 1]$, and none is $0$ or neg­a­tive.

Answer#

(a) $[5, \infty)$; (b) $[-3, \infty)$; (c) $(0, 1]$.

Com­mon mis­take to avoid#

In (c), includ­ing $0$ in the range. A frac­tion with numer­a­tor $1$ is never $0$.

Ques­tion 6: Alge­bra of func­tions#

The prob­lem#

Let $f(x) = 2x - 3$ and $g(x) = x^2$. Find $f + g$, $fg$ and $\displaystyle \frac{g}{f}$, and state the domain of $\displaystyle \frac{g}{f}$.

Under­stand­ing the prob­lem#

These are the point­wise oper­a­tions of the les­son: add, mul­ti­ply or divide the two out­puts at the same $x$. For the quo­tient, the domain must exclude any $x$ where the denom­i­na­tor $f(x)$ is zero.

The idea#

$(f + g)(x) = f(x) + g(x)$, $(fg)(x) = f(x)g(x)$, $\displaystyle \left(\tfrac{g}{f}\right)(x) = \tfrac{g(x)}{f(x)}$ for $f(x) \ne 0$.

Step-by-step solu­tion#

Step 1. Sum:

$$(f + g)(x) = (2x - 3) + x^2 = x^2 + 2x - 3.$$

Step 2. Prod­uct:

$$(fg)(x) = (2x - 3)x^2 = 2x^3 - 3x^2.$$

Step 3. Quo­tient:

$$\displaystyle \left(\frac{g}{f}\right)(x) = \frac{x^2}{2x - 3}.$$

Step 4. Domain of the quo­tient: exclude where $2x - 3 = 0$, i.e. $\displaystyle x = \tfrac32$.

$$\displaystyle \text{Domain} = \mathbf{R} - \left\{\tfrac{3}{2}\right\}.$$

Check­ing the answer#

At $x = 2$: $f(2) = 1$, $g(2) = 4$. Then $f + g = 5$ and $4 + 4 - 3 = 5$ ✓; $fg = 4$ and $16 - 12 = 4$ ✓; $\displaystyle \tfrac{g}{f} = 4$ and $\displaystyle \tfrac{4}{1} = 4$ ✓.

Answer#

$(f + g)(x) = x^2 + 2x - 3$; $(fg)(x) = 2x^3 - 3x^2$; $\displaystyle \left(\frac{g}{f}\right)(x) = \frac{x^2}{2x - 3}$ with domain $\displaystyle \mathbf{R} - \left\{\tfrac32\right\}$.

Ques­tion 7: Sketch­ing the mod­u­lus and great­est inte­ger graphs#

The prob­lem#

Sketch (a) $f(x) = \lvert x - 2 \rvert$ and (b) $f(x) = [x]$ for $-2 \le x < 3$.

Under­stand­ing the prob­lem#

You must describe the shape pre­cisely: key points, where the graph turns or jumps, and which end­points are included.

The idea#

(a) $\lvert x - 2 \rvert$ is the graph of $\lvert x \rvert$ moved $2$ units right, so it is a V with its cor­ner at $x = 2$. (b) $[x]$ is con­stant between con­sec­u­tive inte­gers and jumps up by $1$ at each inte­ger — a stair­case.

Step-by-step solu­tion#

Part (a)

Step 1. Split by sign: for $x \ge 2$, $f(x) = x - 2$; for $x < 2$, $f(x) = 2 - x$.

Step 2. Make a small table.

Step 3. Draw: plot the ver­tex $(2, 0)$ on the $x$-axis. From it draw a straight ray of slope $1$ going up to the right (through $(3,1)$, $(4,2)$) and a ray of slope $-1$ going up to the left (through $(1,1)$, $(0,2)$ — so it meets the $y$-axis at $2$). The result is a V, sym­met­ric about the line $x = 2$, never below the $x$-axis.

Part (b)

Step 1. Work out $[x]$ on each unit inter­val.

Step 2. Draw five hor­i­zon­tal seg­ments, one for each inter­val, at heights $-2, -1, 0, 1, 2$.

Step 3. Mark the ends: each seg­ment has a filled dot at its left end (the inte­ger is included) and an open cir­cle at its right end (the next inte­ger is not included, because there the value jumps up). For exam­ple, the seg­ment at height $0$ runs from a filled dot at $(0, 0)$ to an open cir­cle at $(1, 0)$. The graph looks like a stair­case ris­ing to the right.

Check­ing the answer#

(a) At $x = 2$, $\lvert 0 \rvert = 0$ — the low­est point. (b) $[2.5] = 2$ and $[-1.5] = -2$: these match the heights of the seg­ments over those points.

Answer#

(a) A V-shaped graph with ver­tex $(2, 0)$, arms of slope $\pm1$, meet­ing the $y$-axis at $(0, 2)$. (b) A stair­case of hor­i­zon­tal unit seg­ments at heights $-2, -1, 0, 1, 2$, each closed at the left end and open at the right.

Ques­tion 8: A func­tion on the inte­gers#

The prob­lem#

$f = \{(x, 3x - 1) : x \in \mathbf{Z}\}$. Find $f(4)$ and the value of $x$ for which $f(x) = 14$.

Under­stand­ing the prob­lem#

The set nota­tion says: each inte­ger $x$ is paired with $3x - 1$, so $f(x) = 3x - 1$ with domain the inte­gers. For the sec­ond part the out­put is given and we must find the input — and it must be an inte­ger.

The idea#

Sub­sti­tute for the first part; solve a lin­ear equa­tion for the sec­ond.

Step-by-step solu­tion#

Step 1. $f(4) = 3(4) - 1 = 11$.

Step 2. Solve $f(x) = 14$.

$$3x - 1 = 14 \;\Longrightarrow\; 3x = 15 \;\Longrightarrow\; x = 5.$$

Step 3. $5$ is an inte­ger, so it is in the domain.

Check­ing the answer#

$f(5) = 15 - 1 = 14$ ✓.

Answer#

$f(4) = 11$; $f(x) = 14$ when $x = 5$.

Ques­tion 9: Is y² = x a func­tion of x?#

The prob­lem#

Is the rela­tion $\{(x, y) : y^2 = x\}$ a func­tion of $x$ on $[0, \infty)$? Give a rea­son.

Under­stand­ing the prob­lem#

For it to be a func­tion of $x$, each $x \ge 0$ must cor­re­spond to exactly one $y$.

The idea#

Test one value of $x$: if some $x$ gives two dif­fer­ent $y$'s, the rela­tion is not a func­tion. (Graph­i­cally this is the ver­ti­cal line test.)

Step-by-step solu­tion#

Step 1. Take $x = 4$. Solve $y^2 = 4$: $y = 2$ or $y = -2$.

Step 2. So both $(4, 2)$ and $(4, -2)$ are in the rela­tion — the input $4$ has two images.

Step 3. Graph­i­cally, $y^2 = x$ is a parabola open­ing to the right; the ver­ti­cal line $x = 4$ meets it at two points, so it fails the ver­ti­cal line test.

Check­ing the answer#

Only $x = 0$ has a sin­gle $y$ ($y = 0$); every $x > 0$ has two. One fail­ure is enough.

Answer#

No. For exam­ple $x = 4$ is paired with both $y = 2$ and $y = -2$, so an input has two images.

Ques­tion 10: A taxi fare as a func­tion#

The prob­lem#

A taxi charges ₹$40$ plus ₹$15$ per km. Write the fare as a func­tion of the dis­tance trav­elled, and find the fare for $12$ km.

Under­stand­ing the prob­lem#

There is a fixed charge of ₹$40$ what­ever the dis­tance, plus ₹$15$ for each kilo­me­tre. Let $d$ be the dis­tance in km and $F(d)$ the fare in rupees.

The idea#

Fare $=$ fixed part $+$ (rate $\times$ dis­tance). This is a lin­ear func­tion.

Step-by-step solu­tion#

Step 1. Build the rule.

$$F(d) = 40 + 15d, \qquad d \ge 0.$$

Step 2. Sub­sti­tute $d = 12$.

$$F(12) = 40 + 15 \times 12 = 40 + 180 = 220.$$

Check­ing the answer#

For $0$ km the fare is ₹$40$ (just the fixed charge) — sen­si­ble. Each extra km adds ₹$15$, and $12 \times 15 = 180$.

Answer#

$F(d) = 40 + 15d$ rupees; the fare for $12$ km is ₹$220$.