Solved Examples on the Evaluation of Functions

$f(3)$
This means that we have to substitute $3$ for $x$

$ f(x) = -3x + 7 \\[3ex] f(3) = -3(3) + 7 \\[3ex] f(3) = -9 + 7 \\[3ex] f(3) = -2 \\[3ex] $ $f(-5)$
This means that we have to substitute $-5$ for $x$

$ f(x) = -3x + 7 \\[3ex] f(-5) = -3(-5) + 7 \\[3ex] f(-5) = 15 + 7 \\[3ex] f(-5) = 22 $

$p(0)$
This means that we have to substitute $0$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(0) = -1 * 0^2 - 5(0) + 8 \\[3ex] p(0) = -1 * 0 - 0 + 8 \\[3ex] p(0) = 0 - 0 + 8 \\[3ex] p(0) = 8 $
$p(7)$
This means that we have to substitute $7$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(7) = -1 * 7^2 - 5(7) + 8 \\[3ex] p(7) = -1 * 49 - 35 + 8 \\[3ex] p(7) = -49 - 35 + 8 \\[3ex] p(7) = -76 $
$p(-7)$
This means that we have to substitute $-7$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(-7) = -1 * (-7)^2 - 5(-7) + 8 \\[3ex] p(-7) = -1 * 49 + 35 + 8 \\[3ex] p(-7) = -49 + 35 + 8 \\[3ex] p(-7) = -6 $
$p(g)$
This means that we have to substitute $g$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(g) = -g^2 - 5g + 8 $

$g(0)$
This means that we have to substitute $0$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(0) = 3 * 0^2 - 3 * 0 + 1 \\[3ex] = 3(0) - 0 + 1 \\[3ex] = 0 - 0 + 1 \\[3ex] = 1 $

$g(-2)$
This means that we have to substitute $-2$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(-2) = 3 * (-2)^2 - 3 * (-2) + 1 \\[3ex] = 3(4) + 6 + 1 \\[3ex] = 12 + 6 + 1 \\[3ex] = 19 $

$g(3)$
This means that we have to substitute $3$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(3) = 3 * (3)^2 - 3 * (3) + 1 \\[3ex] = 3(9) - 9 + 1 \\[3ex] = 27 - 9 + 1 \\[3ex] = 19 $

$g(-x)$
This means that we have to substitute $-x$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(-x) = 3 * (-x)^2 - 3 * (-x) + 1 \\[3ex] = 3 * x^2 + 3x + 1 \\[3ex] = 3x^2 + 3x + 1 $

$g(1-k)$
This means that we have to substitute $(1-k)$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(1-k) = 3 * (1-k)^2 - 3 * (1-k) + 1 \\[3ex] = 3 * (1-k) * (1-k) - 3(1-k) + 1 \\[3ex] = 3[(1-k)(1-k)] - 3 + 3k + 1 \\[3ex] = 3(1 - k - k + k^2) + 3k - 2 \\[3ex] = 3(1 - 2k + k^2) + 3k - 2 \\[3ex] = 3 - 6k + 3k^2 + 3k - 2 \\[3ex] = 3k^2 - 3k + 1 $

$g(3)$
This means that we have to substitute $3$ for $x$

$ g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 3 - \dfrac{3}{4} \\[5ex] = \dfrac{2}{1} - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{8 - 3}{4} \\[5ex] = \dfrac{5}{4} \\[5ex] $ $g(4)$
This means that we have to substitute $4$ for $x$

$ g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 4 - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{32}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{32 - 9}{12} \\[5ex] = \dfrac{23}{12} \\[5ex] $ $g\left(-\dfrac{1}{2}\right)$
This means that we have to substitute $-\dfrac{1}{2}$ for $x$

$ g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(4) = \dfrac{2}{3} * -\dfrac{1}{2} - \dfrac{3}{4} \\[5ex] = -\dfrac{1}{3} - \dfrac{3}{4} \\[5ex] = -\dfrac{4}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{-4 - 9}{12} \\[5ex] = -\dfrac{13}{12} $

$p(3)$
This means that we have to substitute $3$ for $m$

$ p(m) = m^3 \\[3ex] p(3) = 3^3 \\[3ex] = 27 \\[3ex] $ $p(-2)$
This means that we have to substitute $(-2)$ for $m$

$ p(m) = m^3 \\[3ex] p(-2) = (-2)^3 \\[3ex] = -8 \\[3ex] $ $p(-m)$
This means that we have to substitute $(-m)$ for $m$

$ p(m) = m^3 \\[3ex] p(3) = (-m)^3 \\[3ex] = (-m)(-m)(-m) \\[3ex] = -m^3 \\[3ex] $ $p(3y)$
This means that we have to substitute $(3y)$ for $m$

$ p(m) = m^3 \\[3ex] p(3y) = (3y)^3 \\[3ex] = 27y^3 \\[3ex] $ $p(2 + h)$
This means that we have to substitute $(2 + h)$ for $m$

$ p(m) = m^3 \\[3ex] p(2 + h) = (2 + h)^3 \\[3ex] = (2 + h)(2 + h)(2 + h) \\[3ex] = (4 + 2h + 2h + h^2)(2 + h) \\[3ex] = (h^2 + 4h + 4)(2 + h) \\[3ex] = (2h^2 + h^3 + 8h + 4h^2 + 8 + 4h) \\[3ex] = h^3 + 6h^2 + 12h + 8 $

$f\left(\dfrac{13}{5}\right)$
This means that we have to substitute $\dfrac{13}{5}$ for $x$

$ f(x) = \sqrt{5x + 3} \\[3ex] f\left(\dfrac{13}{5}\right) = \sqrt{5 * {\dfrac{13}{5}} + 3} \\[3ex] = \sqrt{13 + 3} \\[3ex] = \sqrt{16} \\[3ex] = 4 $

$f(12)$
This means that we have to substitute $12$ for $x$

$ f(x) = \sqrt{5x + 3} \\[3ex] f(12) = \sqrt{5 * 12 + 3} \\[3ex] = \sqrt{60 + 3} \\[3ex] = \sqrt{63} \\[3ex] = \sqrt{9 * 7} \\[3ex] = \sqrt{9} * \sqrt{7} \\[3ex] = 3 * \sqrt{7} \\[3ex] = 3\sqrt{7} $

$f\left(-\dfrac{2}{5}\right)$
This means that we have to substitute $-\dfrac{2}{5}$ for $x$

$ f(x) = \sqrt{5x + 3} \\[3ex] f\left(-\dfrac{2}{5}\right) = \sqrt{5 * -\dfrac{2}{5} + 3} \\[3ex] = \sqrt{-2 + 3} \\[3ex] = \sqrt{1} \\[3ex] = 1 $

$g(0)$
This means that we have to substitute $0$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[3ex] g(0) = \dfrac{0}{\sqrt{1 - 0^2}} \\[3ex] g(0) = \dfrac{0}{\sqrt{1 - 0}} \\[3ex] g(0) = \dfrac{0}{\sqrt{1}} \\[3ex] g(0) = \dfrac{0}{1} \\[3ex] g(0) = 0 $

$g(-1)$
This means that we have to substitute $-1$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[3ex] g(-1) = \dfrac{-1}{\sqrt{1 - (-1)^2}} \\[3ex] (-1)^2 = -1 * -1 = 1 \\[3ex] g(-1) = \dfrac{-1}{\sqrt{1 - 1}} \\[3ex] g(-1) = \dfrac{-1}{\sqrt{0}} \\[3ex] g(-1) = \dfrac{-1}{0} \\[3ex] g(-1) = undefined $
Can you divide "anything" by "nothing"?
Can you create "anything" out of nothing?
Only GOD can do that!

$g(5)$
This means that we have to substitute $5$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[3ex] g(5) = \dfrac{5}{\sqrt{1 - 5^2}} \\[3ex] g(5) = \dfrac{5}{\sqrt{1 - 25}} \\[3ex] g(5) = \dfrac{5}{\sqrt{-24}} $
$g(5)$ is not a real number.
It is an imaginary number.
All the numbers we shall cover in this topic are only real numbers.

$g\left(\dfrac{2}{3}\right)$
This means that we have to substitute $\dfrac{2}{3}$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[5ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{1 - \left(\dfrac{2}{3}\right)^2}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{1 - \left(\dfrac{2^2}{3^2}\right)}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{1 - \dfrac{4}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{\dfrac{9}{9} - \dfrac{4}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{\dfrac{9 - 4}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{\dfrac{5}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} \div \sqrt{\dfrac{5}{9}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} \div \dfrac{\sqrt{5}}{\sqrt{9}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} \div \dfrac{\sqrt{5}}{3} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} * \dfrac{3}{\sqrt{5}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{\sqrt{5}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{\sqrt{5}} * \dfrac{\sqrt{5}}{\sqrt{5}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2\sqrt{5}}{5} $

$f(a)$
This means that we have to substitute $a$ for $x$

$ f(x) = -4x + 7 \\[3ex] f(a) = -4a + 7 $

$f(a + 2)$
This means that we have to substitute $(a + 2)$ for $x$

$ f(x) = -4x + 7 \\[3ex] f(a + 2) = -4(a + 2) + 7 \\[3ex] = -4a - 8 + 7 \\[3ex] = -4a - 1 $

$f(a + h)$
This means that we have to substitute $(a + h)$ for $x$

$ f(x) = -4x + 7 \\[3ex] f(a + h) = -4(a + h) + 7 \\[3ex] = -4a - 4h + 7 $

$g(6)$
This means that we have to substitute $6$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(6) = \dfrac{6 - 5}{6 + 2} \\[5ex] = \dfrac{1}{8} $

$g(5)$
This means that we have to substitute $5$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(5) = \dfrac{5 - 5}{5 + 2} \\[5ex] = \dfrac{0}{7} \\[3ex] = 0 $

$g(-2)$
This means that we have to substitute $(-2)$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(-2) = \dfrac{-2 - 5}{-2 + 2} \\[5ex] = -\dfrac{7}{0} \\[5ex] = undefined \\[3ex] g(-2) DNE $

$g(-12.75)$
This means that we have to substitute $(-12.75)$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(-12.75) = \dfrac{-12.75 - 5}{-12.75 + 2} \\[5ex] = \dfrac{-17.75}{-10.75} \\[5ex] = 1.651162791 $

$g(x + h)$
This means that we have to substitute $(x + h)$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(x + h) = \dfrac{x + h - 5}{x + h + 2} \\[5ex] = \dfrac{x + h - 5}{x + h + 2} $

$f(-a)$
This means that we have to substitute $(-a)$ for $x$

$ f(x) = x^2 - 3x \\[3ex] f(-a) = (-a)^2 - 3(-a) \\[3ex] = a^2 + 3a \\[3ex] = a(a + 3) $

$f(a - 3)$
This means that we have to substitute $(a - 3)$ for $x$

$ f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = a^2 - 3a - 3a + 9 - 3a + 9 \\[3ex] = a^2 - 9a + 18 \\[3ex] = (a - 3)(a - 6) $
OR
$ f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = (a - 3)(a - 3) - 3(a - 3) \\[3ex] = (a - 3)[(a - 3) - 3] \\[3ex] = (a - 3)[a - 3 - 3] \\[3ex] = (a - 3)(a -6) $

$f(a + h)$
This means that we have to substitute $(a + h)$ for $x$

$ f(x) = x^2 - 3x \\[3ex] f(a + h) = (a + h)^2 - 3(a + h) \\[3ex] = (a + h)(a + h) - 3(a + h) \\[3ex] = (a + h)[(a + h) - 3] \\[3ex] = (a + h)(a + h - 3) $

$ f(x) = 4x^3 + Ax^2 + 7x - 5 \\[3ex] f(2) = 4(2)^3 + A(2)^2 + 7(2) - 5 \\[3ex] f(2) = 4(8) + A(4) + 14 - 5 \\[3ex] f(2) = 32 + 4A + 9 \\[3ex] f(2) = 4A + 41 \\[5ex] But\:\: f(2) = -7 \\[3ex] -7 = 4A + 41 \\[3ex] 4A + 41 = -7 \\[3ex] 4A = -7 - 41 \\[3ex] 4A = -48 \\[3ex] A = -\dfrac{48}{4} \\[3ex] A = -12 $

$f(-a)$
This means that we have to substitute $-a$ for $x$

$ f(x) = -x^2 - 3x - 12 \\[3ex] f(x) = -1 * x^2 - 3x - 12 \\[3ex] f(-a) = -1 * (-a)^2 - 3(-a) - 12 \\[3ex] f(-a) = -1 * a^2 + 3a - 12 \\[3ex] f(-a) = -a^2 + 3a - 12 \\[3ex] f(-a) = -1(a^2 - 3a + 12) \\[3ex] $ $f(-a - 2)$
This means that we have to substitute $-a - 2$ for $x$

$ f(x) = -x^2 - 3x - 12 \\[3ex] f(x) = -1 * x^2 - 3x - 12 \\[3ex] f(-a - 2) = -1 * (-a - 2)^2 - 3(-a - 2) - 12 \\[3ex] (-a - 2)^2 = (-a - 2)(-a - 2) \\[3ex] = a^2 + 2a + 2a + 4 \\[3ex] = a^2 + 4a + 4 \\[3ex] f(-a - 2) = -1(a^2 + 4a + 4) - 3(-a - 2) - 12 \\[3ex] f(-a - 2) = -a^2 - 4a - 4 + 3a + 6 - 12 \\[3ex] f(-a - 2) = -a^2 - a - 10 \\[3ex] f(-a - 2) = -1(a^2 + a + 10) $

$f(a + 7)$
This means that we have to substitute $a + 7$ for $x$

$ f(x) = -x^2 - 3x - 12 \\[3ex] f(x) = -1 * x^2 - 3x - 12 \\[3ex] f(a + 7) = -1 * (a + 7)^2 - 3(a + 7) - 12 \\[3ex] (a + 7)^2 = (a + 7)(a + 7) \\[3ex] = a^2 + 7a + 7a + 49 \\[3ex] = a^2 + 14a + 49 \\[3ex] f(a + 7) = -1(a^2 + 14a + 49) - 3(a + 7) - 12 \\[3ex] f(a + 7) = -a^2 - 14a - 49 - 3a - 21 - 12 \\[3ex] f(a + 7) = -a^2 - 17a - 82 \\[3ex] f(a + 7) = -1(a^2 + 17a + 82) $

$ (a.) \\[3ex] g(n) = n^3 - 5n^2 \\[3ex] g(-4n) = (-4n)^3 - 5(-4n)^2 \\[3ex] g(-4n) = -64n^3 - 5(16n^2) \\[3ex] g(-4n) = -64n^3 - 80n^2 \\[3ex] g(-4n) = -16n^2(4n + 5) \\[5ex] (b.) \\[3ex] p(a) = a^3 - 5 \\[3ex] p(x - 4) = (x - 4)^3 - 5 \\[3ex] p(x - 4) = [(x - 4)(x - 4)(x - 4)] - 5 \\[3ex] p(x - 4) = [(x^2 - 4x - 4x + 16)(x - 4)] - 5 \\[3ex] p(x - 4) = [(x^2 - 8x + 16)(x - 4)] - 5 \\[3ex] p(x - 4) = (x^3 - 4x^2 - 8x^2 + 32x + 16x - 64) - 5 \\[3ex] p(x - 4) = (x^3 - 12x^2 + 48x - 64) - 5 \\[3ex] p(x - 4) = x^3 - 12x^2 + 48x - 64 - 5 \\[3ex] p(x - 4) = x^3 - 12x^2 + 48x - 69 \\[5ex] (c.) \\[3ex] k(a) = -4^{3a + 2} \\[3ex] k(a - 2) = -4^{3(a - 2) + 2} \\[3ex] k(a - 2) = -4^{3a - 6 + 2} \\[3ex] k(a - 2) = -4^{3a - 4} $

$f(3, 2)$
We have to substitute $3$ for $x$ and $2$ for $y$

$ f(x, y) = 3x^2 - 4y \\[3ex] f(3, 2) = 3(3)^2 - 4(2) \\[3ex] = 3(9) - 8 \\[3ex] = 27 - 8 \\[3ex] = 19 $

$ g(x) = -[f(x) - 3] \\[3ex] For\:\:g(4),\:\:x = 4 \\[3ex] g(4) = -[f(4) - 3] \\[3ex] f(4) = -1 \\[3ex] \implies \\[3ex] g(4) = -[-1 - 3] \\[3ex] g(4) = -(-4) \\[3ex] g(4) = 4 $

$f(10, 3)$
This means that we have to substitute $10$ for $x$ and $3$ for $y$

$ f(x, y) = xy - (x - y) \\[3ex] f(10, 3) = 10(3) - (10 - 3) \\[3ex] f(10, 3) = 30 - 7 \\[3ex] f(10, 3) = 23 $

$ f(x) = \dfrac{1}{x - 1} + \dfrac{x - 1}{x^2 - 1} \\[5ex] f(1 - x) \\[3ex] = \dfrac{1}{(1 - x) - 1} + \dfrac{(1 - x) - 1}{(1 - x)^2 - 1} \\[5ex] = \dfrac{1}{1 - x - 1} + \dfrac{1 - x - 1}{[(1 - x)(1 - x)] - 1} \\[5ex] = \dfrac{1}{-x} + \dfrac{-x}{(1 - x - x + x^2) - 1} \\[5ex] = -\dfrac{1}{x} + -\dfrac{x}{1 - 2x + x^2 - 1} \\[5ex] = -\dfrac{1}{x} - \dfrac{x}{-2x + x^2} \\[5ex] = -\dfrac{1}{x} - \dfrac{x}{x^2 - 2x} \\[5ex] = -\dfrac{1}{x} - \dfrac{x}{x(x - 2)} \\[5ex] = -\dfrac{1}{x} - \dfrac{1}{(x - 2)} $

$ f(x) = x^2 - 2 \\[3ex] f(x + h) \implies x = x + h \\[3ex] f(x + h) = (x + h)^2 - 2 \\[3ex] (x + h)^2 = (x + h)(x + h) \\[3ex] = x^2 + xh + xh + h^2 \\[3ex] = x^2 + 2xh + h^2 \\[3ex] (x + h)^2 - 2 = x^2 + 2xh + h^2 - 2 \\[3ex] \therefore f(x + h) = x^2 + 2xh + h^2 - 2 $

$ y = 0.005x^2 - 2x + 200 \\[3ex] x = 20 \\[3ex] y = 0.005(20)^2 - 2(20) + 200 \\[3ex] y = 0.005(400) - 40 + 200 \\[3ex] y = 2 - 40 + 200 \\[3ex] y = 162 $

$ f(x + 2) = 2x^2 + 7x - 5 \\[3ex] f(-1) = ? \\[3ex] \implies x + 2 = -1 \\[3ex] x + 2 = -1 \\[3ex] x = -1 - 2 \\[3ex] x = -3 \\[3ex] f(-3) = 2(-3)^2 + 7(-3) - 5 \\[3ex] f(-3) = 2(9) - 21 - 5 \\[3ex] f(-3) = 18 - 21 - 5 \\[3ex] f(-3) = -8 \\[3ex] \therefore f(-1) = -8 $

$ y = \dfrac{x}{(x - 3)} + \dfrac{x}{(x + 4)} \\[5ex] x = -2 \\[3ex] y = \dfrac{-2}{(-2 - 3)} + \dfrac{-2}{(-2 + 4)} \\[5ex] y = \dfrac{-2}{-5} + \dfrac{-2}{2} \\[5ex] y = \dfrac{2}{5} + -1 \\[5ex] y = \dfrac{2}{5} - 1 \\[5ex] y = \dfrac{2}{5} - \dfrac{5}{5} \\[5ex] y = \dfrac{2 - 5}{5} \\[5ex] y = -\dfrac{3}{5} $

$ f(x - 2) = 4x^2 + x + 7 \\[3ex] f(1) = ? \\[3ex] \implies x - 2 = 1 \\[3ex] x - 2 = 1 \\[3ex] x = 1 + 2 \\[3ex] x = 3 \\[3ex] f(3) = 4(3)^2 + 3 + 7 \\[3ex] f(3) = 4(9) + 3 + 7 \\[3ex] f(3) = 36 + 3 + 7 \\[3ex] f(3) = 46 \\[3ex] \therefore f(1) = 46 $

$ f(x) = 2x^2 - 5x + 3 \\[3ex] f(x + 1) \implies x = x + 1 \\[3ex] f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3 \\[3ex] (x + 1)^2 = (x + 1)(x + 1) \\[3ex] = x^2 + x + x + 1 \\[3ex] = x^2 + 2x + 1 \\[3ex] 2(x + 1)^2 = 2(x^2 + 2x + 1) \\[3ex] = 2x^2 + 4x + 2 \\[3ex] 5(x + 1) = 5x + 5 \\[3ex] \implies \\[3ex] f(x + 1) = 2x^2 + 4x + 2 -(5x + 5) + 3 \\[3ex] f(x - 1) = 2x^2 + 4x + 2 - 5x - 5 + 3 \\[3ex] f(x + 1) = 2x^2 - x $

$f(3)$
This means that we have to substitute $4$ for $x$ and $3$ for $y$

$ f(x) = 5x^2 - 7(4x + 3) \\[3ex] f(x) = 5 * x^2 - 7(4 * x + 3) \\[3ex] f(3) = 5 * 3^2 - 7(4 * 3 + 3) \\[3ex] = 5 * 9 - 7(12 + 3) \\[3ex] = 45 - 7(15) \\[3ex] = 45 - 105 \\[3ex] = -60 $

$ f(x) = 3x^2 + 2x - 2 \\[3ex] (a.) \\[3ex] f(0) = 3(0)^2 + 2(0) - 2 \\[3ex] = 3(0) + 0 - 2 \\[3ex] = 0 + 0 - 2 \\[3ex] = -2 \\[5ex] (b.) \\[3ex] f(1) = 3(1)^2 + 2(1) - 2 \\[3ex] = 3(1) + 2 - 2 \\[3ex] = 3 + 2 - 2 \\[3ex] = 3 \\[5ex] (c.) \\[3ex] f(-1) = 3(-1)^2 + 2(-1) - 2 \\[3ex] = 3(1) - 2 - 2 \\[3ex] = 3 - 2 - 2 \\[3ex] = -1 \\[5ex] (d.) \\[3ex] f(-x) = 3(-x)^2 + 2(-x) - 2 \\[3ex] = 3(-x)(-x) - 2x - 2 \\[3ex] = 3x^2 - 2x - 2 \\[5ex] (e.) \\[3ex] -f(x) = -(3x^2 + 2x - 2) \\[3ex] = -3x^2 - 2x + 2 \\[3ex] (f.) \\[3ex] f(x + 2) = 3(x + 2)^2 + 2(x + 2) - 2 \\[3ex] = 3[(x + 2)(x + 2)] + 2x + 4 - 2 \\[3ex] = 3(x^2 + 2x + 2x + 4) + 2x + 2 \\[3ex] = 3(x^2 + 4x + 4) + 2x + 2 \\[3ex] = 3x^2 + 12x + 12 + 2x + 2 \\[3ex] = 3x^2 + 14x + 14 \\[5ex] (g.) \\[3ex] f(4x) = 3(4x)^2 + 2(4x) - 2 \\[3ex] = 3(4^2 * x^2) + 8x - 2 \\[3ex] = 3(16x^2) + 8x - 2 \\[3ex] = 48x^2 + 8x - 2 \\[3ex] = 2(24x^2 + 4x - 1) \\[5ex] (h.) \\[3ex] f(x + h) = 3(x + h)^2 + 2(x + h) - 2 \\[3ex] = 3[(x + h)(x + h)] + 2x + 2h - 2 \\[3ex] = 3(x^2 + hx + hx + h^2) + 2x + 2h - 2 \\[3ex] = 3(x^2 + 2hx + h^2) + 2x + 2h - 2 \\[3ex] = 3x^2 + 6hx + 3h^2 + 2x + 2h - 2 $

$f(4, 3)$
This means that we have to substitute $4$ for $x$ and $3$ for $y$

$ f(x, y) = 3x^2 - 4y \\[3ex] f(4, 3) = 3(4)^2 - 4(3) \\[3ex] f(4, 3) = 3(16) - 12 \\[3ex] f(4, 3) = 48 - 12 \\[3ex] f(4,3) = 36 $