Linear, quadratic and polynomial functions

(a) No. The price $P = 10$ occurs twice, on Tuesday with $D = 55$ and on Thursday with $D = 61.$ A function of $P$ must give the input $10$ exactly one value, and here it would need two. In the plane: the vertical line $P = 10$ meets the set of points twice.

(b) No. For $x = 4$ both $y = 2$ and $y = -2$ satisfy $y^2 = x,$ so the vertical line $x = 4$ meets the set twice. (The set is a parabola lying on its side.)

(c) Yes. For each real number $y$ there is exactly one $x,$ namely $x = y^2.$ So $x$ is a function of $y,$ with domain all real numbers. Whether a curve is the graph of a function depends on which variable is the input.

(a) The slope, then the line. The two rides are the points $(5, 95)$ and $(10, 155).$ By (1.1) the slope is $$a = \frac{155 - 95}{10 - 5} = \frac{60}{5} = 12\,,$$ and by (1.2), $C - 95 = 12(x - 5),$ that is, $$C = 12x + 35\,.$$ Check: at $x = 10$ this gives $120 + 35 = 155.$

(b) Reading the two numbers. The slope $12$ is the price of one more kilometre: $12$ kronor per kilometre. The intercept $35$ is the cost at $x = 0,$ the fixed fee, in kronor. The two numbers have different units, which is one way to tell them apart.

(c) Comparing two lines. The second firm charges $15x.$ It is cheaper when $$ \begin{aligned} &15x < 12x + 35 \\ &\iff 3x < 35 \\ &\iff x < \tfrac{35}{3} \approx 11.7\,. \end{aligned} $$ The two lines have different slopes, so by Fact 1.7 they cross exactly once, at $x = 35/3;$ before that the line without a fixed fee lies lower.

(a) Take out $3,$ then complete. $$ \begin{aligned} &3x^2 - 12x + 7 \\ &\quad= 3\,(x^2 - 4x) + 7 \\ &\quad= 3\,\bigl((x - 2)^2 - 4\bigr) + 7 \\ &\quad= 3\,(x - 2)^2 - 5\,. \end{aligned} $$ Check with (1.4): $-b/2a = 12/6 = 2$ and $-(b^2 - 4ac)/4a = -(144 - 84)/12 = -5.$

(b) A square is never negative. $(x - 2)^2 \ge 0$ for every $x,$ and it is multiplied by the positive number $3,$ so $f(x) = 3(x - 2)^2 - 5 \ge -5.$ There is equality exactly when the bracket is zero, that is, at $x = 2.$ So $-5$ is the smallest value of $f,$ and $f$ has no largest value.

(c) Move the parabola up. In the same way $3x^2 - 12x + c = 3(x - 2)^2 + (c - 12).$ The first term is at least $0,$ so the expression is never zero when $c - 12 > 0.$ If $c = 12$ there is one root, $x = 2,$ and if $c < 12$ there are two. The discriminant says the same: $\Delta = 144 - 12c < 0$ exactly when $c > 12.$

(a) The demand line. Five kronor more loses ten customers, so the slope is $-10/5 = -2$ tickets per krona. The line passes through $(p, q) = (100, 300),$ so by (1.2) $$q - 300 = -2(p - 100)\,,$$ that is, $q = 500 - 2p.$

(b) Revenue is a quadratic in $p.$ $$ \begin{aligned} R(p) = p\,q &= p\,(500 - 2p) \\ &= 500p - 2p^2\,. \end{aligned} $$ Here $a = -2 < 0,$ so by Fact 1.11 there is a maximum, at $$p^{*} = -\frac{500}{2\cdot(-2)} = 125\,.$$ The largest revenue is the value there: $R(125) = 125\cdot 250 = 31\,250$ kronor. (At the present price the revenue is $100\cdot 300 = 30\,000.$)

(c) The constraints. The price must not be negative, the number of tickets must not be negative, and it must not exceed the $400$ seats. At $p = 125$ the cinema sells $q = 500 - 250 = 250$ tickets, and $0 \le 250 \le 400,$ so the answer is allowed. (Had the vertex been at a price below $50,$ the formula would have asked for more than $400$ tickets, and the answer would have had to be changed.)

(a) The firm's problem. With the tax the cost is $12Q + Q^2 + \tau Q,$ so the profit $\pi(Q)$ is $$ \begin{aligned} &(60 - Q)Q - 12Q - Q^2 - \tau Q \\ &\quad= (48 - \tau)Q - 2Q^2\,. \end{aligned} $$ This is a quadratic in $Q$ with $a = -2 < 0$ and $b = 48 - \tau,$ so the maximum is at $$Q^{*}(\tau) = -\frac{48 - \tau}{2\cdot(-2)} = \frac{48 - \tau}{4}\,.$$ For $0 \le \tau \le 48$ this is $\ge 0,$ so the constraint $Q \ge 0$ holds.

(b) The government's problem. Now $\tau$ is the variable: $$ \begin{aligned} T(\tau) &= \tau\cdot\frac{48 - \tau}{4} \\ &= 12\tau - \tfrac14\tau^2\,. \end{aligned} $$ A quadratic in $\tau$ with $a = -\tfrac14 < 0$ and $b = 12,$ so it has a maximum, at $$\tau^{*} = -\frac{12}{2\cdot(-\tfrac14)} = 24\,,$$ and $24$ lies in the allowed interval $[0, 48].$ The largest tax revenue is the value there: $T(24) = 24\cdot 6 = 144.$

(c) Reading the answer. At $\tau = 24$ the firm produces $Q^{*} = (48 - 24)/4 = 6$ units; with no tax it produces $48/4 = 12.$ The revenue-maximising tax halves the output. At $\tau = 48$ the output is $0$: the firm shuts down, and the tax brings in nothing. The roots of $T$ are $\tau = 0$ and $\tau = 48,$ and its vertex lies midway between them, at $24.$

Integer coefficients first (Method 1.19, step 1). Multiply the equation by $2;$ this does not change the roots: $$2x^3 - x^2 - 8x + 4 = 0\,.$$

The candidates (Fact 1.18). An integer root must divide the constant term $4,$ so the candidates are $\pm1, \pm2, \pm4.$

The test. Write $P(x) = 2x^3 - x^2 - 8x + 4.$ Then $$ \begin{aligned} P(1) &= 2 - 1 - 8 + 4 = -3, \\ P(-1) &= -2 - 1 + 8 + 4 = 9, \\ P(2) &= 16 - 4 - 16 + 4 = 0, \\ P(-2) &= -16 - 4 + 16 + 4 \\ &= 0, \\ P(4) &= 128 - 16 - 32 + 4 \\ &= 84, \\ P(-4) &= -128 - 16 + 32 + 4 \\ &= -108. \end{aligned} $$ So $2$ and $-2$ are roots, and they are the only integer roots.

Divide the factors out (Fact 1.17). $x - 2$ and $x + 2$ are factors, and so is their product $x^2 - 4.$ Dividing, $$ \begin{aligned} &2x^3 - x^2 - 8x + 4 \\ &\quad= (x^2 - 4)(2x - 1)\,. \end{aligned} $$ Check: $(x^2 - 4)(2x - 1) = 2x^3 - x^2 - 8x + 4.$

The root the test could not find. The remaining factor $2x - 1$ is zero at $x = \tfrac12,$ which is not an integer. This is Watch out 1.21: “no further integer roots” is not “no further roots”. A cubic has at most three roots, and we have three, so there are no more.

(a) The numerator has degree $3$ and the denominator degree $2,$ so $R$ is improper (Definition 1.24). It is defined wherever $x^2 - 1 \ne 0,$ that is, for all $x$ except $1$ and $-1.$

(b) Divide. $2x\cdot x^2 = 2x^3,$ so the first term of the quotient is $2x,$ and $$ \begin{aligned} &(2x^3 + x^2 - 5) - 2x(x^2 - 1) \\ &\quad= x^2 + 2x - 5\,. \end{aligned} $$ Next $1\cdot x^2 = x^2,$ so the second term is $1,$ and $$ \begin{aligned} &(x^2 + 2x - 5) - (x^2 - 1) \\ &\quad= 2x - 4\,. \end{aligned} $$ The remainder $2x - 4$ has degree $1 < 2,$ so we stop: $$R(x) = 2x + 1 + \frac{2x - 4}{x^2 - 1}\,.$$

The check. $(2x + 1)(x^2 - 1) = 2x^3 + x^2 - 2x - 1,$ and adding the remainder $2x - 4$ gives $2x^3 + x^2 - 5,$ the numerator.

(c) At $x = 10$ the polynomial part is $2\cdot 10 + 1 = 21$ and the proper part is $\dfrac{16}{99} \approx 0.16.$ Together $R(10) = \dfrac{2095}{99} \approx 21.16.$ The polynomial part carries nearly everything: for large $|x|$ the proper part dies away, and the graph of $R$ follows the line $y = 2x + 1.$