(a)(i) \(f(-2)=14\), \(f(0)=10\), \(f(3)=4\). The range is \(\{14,10,4\}\).

(a)(ii) From \(y=10-2x\), rearrange and interchange variables: \(f^{-1}(x)=\frac{10-x}{2}\).

(a)(iii) \(gf(x)=g(f(x))=2(10-2x)^2+3=8x^2-80x+203\).

(b) Solve \(10^{2/x}=10000=10^4\). Thus \(2/x=4\), giving \(h^{-1}(10000)=\frac12\).

At \(x=0\), \(80=ab^0=a\). At \(x=30\), \(20=80b^{30}\), so \(b^{30}=1/4\), and \(b=(1/4)^{1/30}=0.9548416\ldots\approx0.955\).

(a)(i) \(f(7)=49-1=\boxed{48}\).

(a)(ii) \((10-3x)^2-1=120\), so \((10-3x)^2=121\) and \(10-3x=\pm11\). This gives \(x=-1/3\) or 7; the positive solution is \(\boxed{7}\).

(b) \(h(-5)=-13,h(0)=-3,h(2)=1\), so the range is \(\boxed{\{-13,-3,1\}}\).

(c) Applying j, \(x=j(4)=2^{12}=\boxed{4096}\).

(a) \(f(4)=20-12=\boxed{8}\).

(b) From \(y=7+2x\), \(x=(y-7)/2\). Hence \(\boxed{h^{-1}(x)=\frac{x-7}{2}}\).

(c) \(g(5)=64/32=2\), so \(gg(5)=g(2)=64/4=\boxed{16}\).

(d) Apply g to both sides: \(x=g(8)=64/256=\boxed{\frac14}\).

(a)(i) \(3x+1=5-3x\), so \(6x=4\) and \(\boxed{x=2/3}\).

(a)(ii) From \(y=5-3x\), \(x=(5-y)/3\), giving \(\boxed{g^{-1}(x)=\frac{5-x}{3}}\).

(a)(iii) \(g(g(x))=5-3(5-3x)=\boxed{9x-10}\).

(b) Apply h to both sides: \(x=h(2)=2^4=\boxed{16}\).

(a) \(5-4(-3)=\boxed{17}\).

(b) \(5-4(3-2x)=\boxed{8x-7}\).

(c) Rearrange \(y=5-4x\) to \(x=(5-y)/4\). Hence \(\boxed{f^{-1}(x)=\frac{5-x}{4}}\).

(a) \(f(-2)=3(-2)-4=\boxed{-10}\).

(b) Rearrange \(y=3x-4\): \(x=(y+4)/3\). Thus \(\boxed{f^{-1}(x)=(x+4)/3}\).

(c) \(f(g(x))=3(4x+1)-4=12x-1\), so \(\boxed{a=12,b=-1}\).

(d) \(\frac2{3x-4}-\frac5{4x+1}=\frac{2(4x+1)-5(3x-4)}{(3x-4)(4x+1)}=\boxed{\frac{22-7x}{(3x-4)(4x+1)}}\).

(a) \(2(3)+5=\boxed{11}\).

(b) Rearrange \(y=2x+5\), giving \(\boxed{f^{-1}(x)=(x-5)/2}\).

(c) \(f(g(x))=2(x-4)+5=2x-3\). Thus \(2x-3=25\), so \(\boxed{x=14}\).

(d) Apply h to both sides: \(x=h(2)=5^2=\boxed{25}\).

(a) \(-3+1=\boxed{-2}\).

(b) \(g(-3)=11,g(0)=5,g(2)=1\), giving \(\boxed{\{11,5,1\}}\).

(c) \(1/32=2^{-5}\), so \(\boxed{x=-5}\).

(d) Apply h: \(x=h(3)=\boxed{8}\).

\(ff(x)=f(2x+5)=2(2x+5)+5=4x+15\). Therefore \((2x+5)^2-(4x+15)=4x^2+20x+25-4x-15=4x^2+16x+10\). Comparing coefficients gives \(\boxed{a=4,b=16,c=10}\).

(a)(i) \(g(2)=6+4=\boxed{10}\). (ii) \(f(-1)=21/(-3)=-7\), so \(g(-7)=-21+4=\boxed{-17}\).

(iii) Write \(y=21/(2x-1)\). Then \(2xy-y=21\), so \(x=(21+y)/(2y)\). Interchange x and y: \(\boxed{f^{-1}(x)=(21+x)/(2x)}\), \(x\ne0\).

(b) \(21=(3x+4)(2x-1)=6x^2+5x-4\). Thus \(6x^2+5x-25=(3x-5)(2x+5)=0\), giving \(\boxed{x=5/3\text{ or }-5/2}\). Neither root is excluded.

(a) \(7^{x-4}=7^0\), so \(\boxed{x=4}\). (b) Apply f to both sides: \(x=f(1)=7^{-3}=\boxed{1/343}\).

(a) \(\boxed{5^5=3125}\). (b) \(3(8x)-2=\boxed{24x-2}\). (c) Reverse y=3x−2 to give \(\boxed{g^{-1}(x)=(x+2)/3}\).

(d) \(g(h(x))=3(x^2+1)-2=3x^2+1=364\), so x²=121. The positive root is \(\boxed{11}\). (e) A function composed with its inverse returns the input: \(\boxed{12}\).

(a)(i) f(2)=1/2, so \(g(f(2))=3/2-5=\boxed{-3.5}\). (ii) Reverse y=3x−5 to give \(\boxed{g^{-1}(x)=(x+5)/3}\). (b) \(3(x-2)-5=\boxed{3x-11}\).

(c)(i) \(1/(3x-5)=0.1\), so 3x−5=10, giving \(\boxed{x=5}\). (ii) g(7)=16, so \(2^x=16=2^4\), giving \(\boxed{x=4}\).

(a) \(3^x=243=3^5\), so \(\boxed{x=5}\). (b) Apply f to both sides: \(x=f(7)=3^7+2=\boxed{2189}\).

(a)(i) 4×3+1=\(\boxed{13}\). (ii) g(13)=6−26=\(\boxed{-20}\). (b) Solve y=6−2x for x: \(\boxed{g^{-1}(x)=(6-x)/2}\). (c) \(4x+1=6-2(2x-7)=20-4x\), so 8x=19 and \(\boxed{x=19/8=2.375}\). (d) h(2)=1; h(1)=\(3^{-1}=\boxed{1/3}\). (e) \(x=h(10)=3^8=\boxed{6561}\).

(a) 7x=14 ⇒\(\boxed{x=2}\). (b) \(\boxed{5a(2a+1)}\). (c) Difference of squares: \((2x-6)(2x)=\boxed{4x(x-3)}\). (d)(i) \(\boxed{1/15}\). (ii) g(2)=9, g(9)=3⁹=\(\boxed{19683}\). (iii) \(3^k=f(7)=1/27=3^{-3}\), so \(\boxed{k=-3}\).

(a) Since 4⁰=1, \(x+3=0\), so \(\boxed{x=-3}\). (b) \(1/16=4^{-2}\). Solve \(x+3=-2\): \(\boxed{g^{-1}(1/16)=-5}\).

(a)(i) x²=36, so \(\boxed{x=\pm6}\). (ii) y=7−3x gives \(\boxed{f^{-1}(x)=(7-x)/3}\). (iii) \((7-3x)^2-16+1=\boxed{9x^2-42x+34}\). (iv) Upward parabola with intercepts (−4,0),(4,0),(0,−16), the last being its minimum. (v) g′(x)=2x, so slope=−6 at (−3,−7). \(y+7=-6(x+3)\), hence \(\boxed{y=-6x-25}\). (b)(i) Increasing exponential through (0,1), approaching zero to the left. (ii) \(\boxed{y=0}\).

(a)(i) \(2(7)-5=\boxed{9}\). (ii) \(g(9)=81-18=\boxed{63}\). (iii) y=2x−5 gives \(\boxed{f^{-1}(x)=(x+5)/2}\). (b) \((2x-5)^2-2(2x-5)-3(x^2-2x)=4x^2-20x+25-4x+10-3x^2+6x=\boxed{x^2-18x+35}\).

(a) \(1-2(-3)=\boxed{7}\). (b) \((2x+5)(1-2x)+[2(1-2x)+5]+1=-4x^2-8x+5+7-4x+1=\boxed{-4x^2-12x+13}\). (c) y=1−2x givesx=(1−y)/2, so \(\boxed{g^{-1}(x)=(1-x)/2}\). (d) h(1)=1/2, then \(h(1/2)=\boxed{2/3}\). (e) \(\frac1{2x+5}-\frac1{x+1}=\frac{x+1-(2x+5)}{(2x+5)(x+1)}=\boxed{\frac{-x-4}{(2x+5)(x+1)}}\), forx≠−5/2,−1. (f) \(2^x=2^{-5}\), so \(\boxed{x=-5}\). (g) \(j^{-1}(x)=0\) meansx=j(0), so \(\boxed{x=1}\).

Set \(y=x^3+1\). Rearranging gives \(x=\sqrt[3]{y-1}\). Therefore \(f^{-1}(x)=\sqrt[3]{x-1}\).

(a) \(j(-1)=(-1)^2=1\). (b) \(2x-1+3x+2=0\Rightarrow x=-1/5\).

(c) \(g(g(x))=3(3x+2)+2=9x+8\).

(d) \(\frac1{2x-1}+\frac3x+2=\frac{x+3(2x-1)+2x(2x-1)}{x(2x-1)}=\frac{4x^2+5x-3}{x(2x-1)}\), \(x\ne0,1/2\).

(e) \(h\), since \(h(h(x))=1/(1/x)=x\).

\(y=5x+2\Rightarrow x=(y-2)/5\). Hence \(f^{-1}(x)=\frac{x-2}5\).

(a) \(6(x+2)-7=6x+5\). (b) \(y=6x-7\Rightarrow x=(y+7)/6\), so \(f^{-1}(x)=(x+7)/6\).

(c) \(f(22)=125\). \(x^{-3}=125\Rightarrow x^3=1/125\Rightarrow x=1/5\).

(a)(i) \(-3-4=-7\). (ii) \(g^{-1}(x)=\frac{x-5}2\).

(iii) \((x-4)^2(2x+5)=(x^2-8x+16)(2x+5)=2x^3-11x^2-8x+80\).

(b) \(f(2)=-2,\quad g(-2)=1\). Thus \(3^x=1\Rightarrow x=0\).

(a)(i) \((1)(5)(-4)=-20\). (ii) \(g^{-1}(x)=\frac{x+3}2\). (iii) \(h(2)=4^3=64\), \(g(64)=128-3=125\).

(b) \(4x-3=7\Rightarrow x=2.5\). (c) \((2x^2-3)+(4x-9)+1=2x^2+4x-11\).

(d) Solve \(4^{2x-1}=16=4^2\): \(2x-1=2\), so \(x=1.5\). (e) \((x+5)(x-4)=x^2+x-20\). Multiply by \(3x+1\) to get \(3x^3+4x^2-59x-20\). Thus \(a=3,b=4,c=-59,d=-20\).

(a) \((3-1)^2=4\). (b) \(1-3(x-2)=7-3x\). (c) \(f^{-1}(x)=\frac{1-x}3\).

(d) \(g(f(x))=(-3x)^2=9x^2\). \(g(x)f(x)=(x^2-2x+1)(1-3x)=-3x^3+7x^2-5x+1\). Subtract to get \(3x^3+2x^2+5x-1\), so \(a=2,b=5,c=-1\).

(e) \(\frac3x-(1-3x)=\frac{3-x+3x^2}x\). (f) \(\frac3{3x^7}=x^{-7}\Rightarrow n=-7\).

(a)(i) \(5(2)-3=7\). (ii) \(f(0.5)=-0.5\), so \(64^{-1/2}=1/8\).

(b) \(y=\frac2{x+1}\Rightarrow x=\frac2y-1\). Hence \(h^{-1}(x)=\frac2x-1,\quad x\ne0\). (c) \(2^{6x}=2^{-5}\Rightarrow x=-5/6\).

(d) \(\frac1{5x-3}-\frac2{x+1}=\frac{x+1-2(5x-3)}{(5x-3)(x+1)}=\frac{7-9x}{(5x-3)(x+1)}\), with \(x\ne3/5,-1\).

(a)(i) \(f(3)=13\), so \(g(13)=169\). (ii) \(1+4x^2\). (iii) \(x\), since the inverse undoes \(f\). (b) \(1+4x=15\) gives \(x=3.5\).

(a) \(f^{-1}(x)=\frac{x+8}7\). (b) \(g(1/3)=12+5=17\). \(2^x+1=17\) gives \(2^x=16\), hence \(x=4\).

(a) \(k(-5k)^2=25k^3=675\), so \(k^3=27\) and \(k=3\). (b) \(\frac5{7x-2}\).

(c) \(h^{-1}(x)=\frac{5x+2}7\). Therefore \(\frac{5x+2}7+\frac{3-10x}{14}=\frac{10x+4+3-10x}{14}=\frac12\).

(a)(i) \(3\). (ii) \(g(3)=7\). (b) \(g^{-1}(x)=(x+2)/3\). (c) \(1/x=5^{-2}=1/25\) gives \(x=25\).

(d) \(2x-1-1/x=\frac{2x^2-x-1}{x}\). (e) \(5^{25}=2.98\times10^{17}\). (f) \(x=j(4)=5^4=625\).

(a) \(f(-3)=9\). (b) \(y=(x+5)/2\) gives \(x=2y-5\), hence \(g^{-1}(x)=2x-5\). (c) \(h(h^{-1}(63))=63\). Thus \((x^2+5)/2=63\), \(x^2=121\). As x > 0, \(x=11\).

(a) \(g(f(x))=\frac{10}{5x-5}=\frac2{x-1}\). (b) \(y=10/(x-2)\) gives \(x=10/y+2\). Therefore \(g^{-1}(x)=\frac{10}x+2=\frac{2x+10}x\), \(x\ne0\). (c) An inverse followed by f returns its input, giving \(x-1\) (where the composition is defined, x > 3).

(a)(i) \(2/(1/2)=4\). (ii) \(2^4=16\). (b) \(10-x=7\) gives x = 3. (c) \(2/x=8\) gives \(x=1/4\). (d) \(y=5-2x\) gives \(j^{-1}(x)=(5-x)/2\).

(e) \(11-x+2/x=\frac{-x^2+11x+2}x\). (f) \(f(f(x))=10-(10-x)=x\). Thus \((10-x)^2-x=x^2-21x+100\) gives \(a=1,b=-21,c=100\). (g) \(x=h(10)=2^{10}=1024\).

\(f(g(x+1))=(x+5)^2-25=x^2+10x\). \(g(f(x))=x^2-21\). Equate: \(10x=-21\), so \(x=-2.1\).

(a) \(3-2(-5)=13\). (b) \(3-2(3-2x)=4x-3\). (c) \(x^2+5=3-2x+37\) gives \(x^2+2x-35=(x+7)(x-5)=0\), so x = −7 or 5. (d) \(y=3-2x\) gives \(x=(3-y)/2\), hence \(f^{-1}(x)=(3-x)/2\). (e) \((3-2x)^3+x^2+5=27-54x+36x^2-8x^3+x^2+5=-8x^3+37x^2-54x+32\).

(a) f(2) = 4, g(2) = 3, gf(2) = g(4) = 13. (b) Rearrange \(y=3x-2\) to get \(f^{-1}(x)=(x+2)/3\). (c) \((3x-2)^2+(3x-2)=9x^2-9x+2\). (d) \(h^\prime(x)=2x+1\). (e)(i) Apply j to both sides: \(x=j(4)=3^4=81\). (ii) A function and its inverse cancel, giving x.

(a) \(f(6)=2^3=8\), so \(f(8)=2^5=32\). (b) An inverse undoes g: \(x+21\). (c) \(h(84)=5/80=1/16=2^{-4}\). Hence \(x-3=-4\) and \(x=-1\).

(a)(i) \(3-5x=-5\) gives \(x=8/5=1.6\). (ii) \(f^{-1}(x)=\frac{3-x}5\). (b)(i) \(18-3x-x^2=20.25-(x+1.5)^2\). (ii) Downward parabola, maximum at (−1.5, 20.25), x-intercepts −6 and 3, y-intercept 18. (iii) \(\frac{dy}{dx}=-3-2x\), so gradient at x = 4 is −11. Point \((4,-10)\). \(y+10=-11(x-4)\) gives \(y=-11x+34\).

(a)(i) \(4^3=64\). (ii) \(2(64)-1=127\). (b) \(g(f(x))=(2x-1)^2+2(2x-1)=4x^2-1=0\). Hence \(x=\pm1/2\). (c) p is the inverse of f: \(p(x)=(x+1)/2\). (d) \(4^x2^x=2^{3x}=2^{-1/2}\), so \(3x=-1/2\) and \(x=-1/6\).

(a)(i) \(4(x-3)-1=4x-13\). (ii) \((5x)^2=25x^2\). (b) \(y=4x-1\) gives \(x=(y+1)/4\), hence \(f^{-1}(x)=(x+1)/4\). (c) \(h(1)=1/3\), so \(h(1/3)=3^{-1/3}=.6934\). (d)(i) \((3x-2)^2-3^3=9x^2-12x+4-27=9x^2-12x-23\). (ii) \(x=\frac{12\pm\sqrt{144+828}}{18}=\frac{12\pm\sqrt{972}}{18}\), giving −1.07 and 2.40. (e) \(f(61)=243=3^5=3^{-x}\), so x = −5.

(a) \(f(g(x))=4(5x-4)+3=20x-13\), so p = −13. (b) \(y=(5x-1)/3\) gives \(3y+1=5x\), hence \(h^{-1}(x)=\frac{3x+1}{5}\).

(a) \(4^4=256\). (b) g(1)=2, then f(2)=8. (c) \(g(f(x))=(3x+2)^2+1=9x^2+12x+5\).

(d) g(7)=50; \(3x+2=50\) gives x=16. (e) \(f^{-1}(x)=(x-2)/3\). (f) \(\frac{x^2+1}{3x+2}+x=\frac{x^2+1+x(3x+2)}{3x+2}=\frac{4x^2+2x+1}{3x+2}\). (g) Apply h to both sides: x = h(2)=16.

(a) \(12/(6-3)=4\). (b) g(4)=8, then f(8)=52. (c) \(f(x^2)=7x^2-4\).

(d) \(\frac{7x-4}2+\frac{2x}{x-3}=\frac{(7x-4)(x-3)+4x}{2(x-3)}=\frac{7x^2-21x+12}{2(x-3)}\). (e) \(7(x+2)-4=-11\), so \(7x=-21\) and x=−3. (f) \(p^2=p\Rightarrow p(p-1)=0\), so p=0 or1.

(a) \(3(-2)^2+a=19\) gives a=7. (b)(i) \(2(3x-8)+7=6x-9\). (ii) Rearranging y=2x+7 gives \(g^{-1}(x)=(x-7)/2\).

(a)(i) \(3^2+1=10\). (ii) \(g(10)=1-20=-19\). (b) Rearrange y=1−2x: \(g^{-1}(x)=(1-x)/2\). (c) 1/x=2 gives x=1/2. (d) \((1-2x)^2-[1-2(1-2x)]=4x^2-8x+2\). (e) \(1/(1/x)=x\), x≠0. (f) \(5^5=3125\). (g) \(x=j(2)=25\). (h) \(g(-12)=25,\ h(25)=1/25=5^{-2}\). Hence x=−2.

(a) h(2)=9, then f(9)=4−27=−23. (b) Rearrange y=4−3x: \(f^{-1}(x)=(4-x)/3\). (c)(i) \(4-3(1-2x)=1+6x\). (ii) \((4-3x)^2+(4-3x)-9(x^2+x)=20-36x\). (d) \(3^{-x}=3^{2kx}\), so −1=2k and k=−1/2.

(a)8/4−5=−3. (b) f(2)=3/4, so \(ff(2)=3/(3/4+2)=12/11\). (c) \(8(8x-5)-5=64x-45\). (d) \(g^{-1}(x)=(x+5)/8\). (e) \(\frac{(8x-5)(x+2)-3}{x+2}=\frac{8x^2+11x-13}{x+2}\), x≠−2.

(f)(i) \((8x-5)^2+6=19\Rightarrow64x^2-80x+12=0\). Divide by4: \(16x^2-20x+3=0\). (ii) \(x=\frac{20\pm\sqrt{(-20)^2-4(16)(3)}}{32}=\frac{20\pm\sqrt{208}}{32}\). Roots0.174306… and1.075694…, so x=0.17 or1.08.

\(f(1-x)=2(1-x)+3=5-2x\).

(a) h(2)=9; g(9)=82. (b) \(f^{-1}(x)=(x+2)/7\). (c) \((x^2+1)^2+1=x^4+2x^2+2\), so a=1,b=2,c=2. (d) \(3^{7x-2}=81=3^4\Rightarrow7x-2=4\Rightarrow x=6/7\).

(a) g(3)=8; f(8)=19. (b) \(y=3x-5\Rightarrow x=(y+5)/3\), so \(f^{-1}(x)=(x+5)/3\).

(a)(i) 7−2(−3)=13. (ii) \(g(30)=\frac13,\quad h(\frac13)=\sqrt[3]{27}=3\). (iii) \(y=7-2x\Rightarrow x=\frac{7-y}2\), so \(f^{-1}(x)=\frac{7-x}2\).

(b) \(\frac{10}{2x+1}=4\Rightarrow10=8x+4\Rightarrow x=.75\). (c) \(\frac1{7-2x}+\frac{10}x=\frac{x+10(7-2x)}{x(7-2x)}=\frac{70-19x}{x(7-2x)}\), x≠0,3.5. (d) \(27^3=19683\), so the inverse value is 3.

(a)(i) f(4)=8−3=5. (ii) g(3)=0, so h(0)=1. (iii) \(9-4x^2\). (iv) \(2(9-x^2)-3=15-2x^2\). (b) \(y=2x-3\Rightarrow x=\frac{y+3}2\), so \(f^{-1}(x)=\frac{x+3}2\).

(c) \(5(2x-3)=3\Rightarrow10x=18\Rightarrow x=1.8\). (d) \(9-(2x-3)^2=-16\Rightarrow(2x-3)^2=25\Rightarrow2x-3=\pm5\), giving x=−1 or 4. (e) \(x=h(-2)=3^{-2}=\frac19\).

(a) f(2)=5, then f(5)=2. (b) \(4(7-x)+2=30-4x\). (c) \(15-(2x)^2=15-4x^2\).

(a) f(−3)=11, then f(11)=−17. (b)(i) \(g(2x)=4x^2+8\). (ii) \(y=5-2x\Rightarrow x=\frac{5-y}2\), so \(f^{-1}(x)=\frac{5-x}2\).

(a)(i) \(f(8/3)=0\Rightarrow h(0)=1\). (ii) \(h(-2)=1/4\Rightarrow g(1/4)=10/(5/4)=8\). (iii) \(y=\frac{10}{x+1}\Rightarrow x=\frac{10}y-1\), so \(g^{-1}(x)=\frac{10}x-1,\ x\ne0\). (iv) 5. (b) \(\frac{(8-3x)(x+1)+10}{x+1}=\frac{-3x^2+5x+18}{x+1}\).

(a)(i) g(2)=5, then g(5)=26. (ii) \((x+2)^2+1=x^2+4x+5\). (iii)2x−3=7 gives x=5. (iv) \(f^{-1}(x)=\frac{x+3}2\). (b)(i) \(.3^{.3}=.696845\ldots\approx.70\). (ii) x=4, because \(4^4=256\).

(a) \(3^{3x}=(3^3)^x=27^x\), so k=27. (b)7+3 x=16 gives x=3. (c) \(f^{-1}(x)=\frac{x-7}3\).

(a)(i) \(5x^3+2\). (ii) \(g^{-1}(x)=\frac{x-2}5\). (b)4 a+1=21 gives a=5.

(a)0. (b)h(−1)=1/3, f(1/3)=5. (c) \(g^{-1}(x)=\frac{x+1}2\). (d)3(3 x+4)+4=9 x+16. (e)(3 x+4)²=9 x²+24 x+16. (f)g(2)=3; apply h to both sides: x=h(3)=27.

(a) \(x/4-3=-0.5\Rightarrow x=10\). (b) \(y=6x-7\Rightarrow x=\frac{y+7}6\), so \(g^{-1}(x)=\frac{x+7}6\). (c) \(f(13)=\frac14=2^{-2}\), so x=−2.

(a) \(3+4(3x)=3+12x\). (b) \(f(g(x))=3+4(6x+7)=24x+31\).

(a) 3(−3)−2=−11. (b) 3x−2=19 gives x=7. (c) h(2)=9, so f(9)=25. (d) \((3x-2)^2+(3x-2)+x=9x^2-8x+2\). (e) \(y=3x-2\Rightarrow x=(y+2)/3\), so \(f^{-1}(x)=(x+2)/3\).

\(g(f(x))=(5x-3)^2+6(5x-3)+1=25x^2-30x+9+30x-18+1=25x^2-8\).

(a) −7. (b) \(h(3)=64,\quad g(64)=4/64=1/16\). (c) \(y=3-2x\Rightarrow x=(3-y)/2\), so \(f^{-1}(x)=(3-x)/2\).

(d) \(4^{3-2x}=4^3/(4^2)^x=64/16^x\). (e) g(0.5)=8; \(4^x=8\Rightarrow2^{2x}=2^3\Rightarrow x=1.5\).

(a) 3. (b) \(2-4x=x+4\Rightarrow x=-2/5\). (c) \(1-2(x+4)=-2x-7\). (d) h(2)=5,h(5)=26. (e) \(y=1-2x\Rightarrow x=(1-y)/2\), hence \(f^{-1}(x)=(1-x)/2\). (f) \((5-2x)^2+1=4x^2-20x+26\), so p=−20,q=26.

(a)(i) \(2-3(-3)=11\). (ii) \(7(2x)+3=14x+3\). (b) \(g(f(x))=7(2-3x)+3=17-21x\). (c) \(3(2-3x)=7\Rightarrow6-9x=7\Rightarrow x=-\frac19\). (d) \(2-3(x+4)-(7x+3)=0\Rightarrow-10x-13=0\Rightarrow x=-1.3\).

(a) \(g(1)=5\), so \(2x+1=5\) and \(x=2\). (b) \(h(3)=8\) and \(f(8)=17\). (c) Rearranging \(y=2x+1\) gives \(x=(y-1)/2\), so \(f^{-1}(x)=\frac{x-1}{2}\). (d) \(g(f(x))=(2x+1)^2+4=4x^2+4x+5\). (e) Apply h to both sides: \(x=2^{0.5}=\sqrt2\approx1.41\). (f) \(\frac1{2^x}=2^{-x}\), hence \(k=-1\).

(a)(i) \(g(1)=4/(1-3)=-2\), so \(f(-2)=-10+7=-3\). (ii) \(g(f(x))=\frac4{5x+7-3}=\frac4{5x+4}\). (iii) From \(y=\frac4{x-3}\), \(y(x-3)=4\Rightarrow x=3+\frac4y\). Therefore \(g^{-1}(x)=3+\frac4x,\quad x\ne0\). (iv) Applying the inverse undoes f, giving 2. (b)(i) \(5x+7=\frac4{x-3}\Rightarrow(5x+7)(x-3)=4\). Expand: \(5x^2-8x-21=4\), hence \(5x^2-8x-25=0\). (ii) \(x=\frac{8\pm\sqrt{64+500}}{10}=\frac{8\pm\sqrt{564}}{10}\). The roots are \(-1.574868\ldots\) and \(3.174868\ldots\), so \(x=-1.57\) or \(x=3.17\).

(a) \(f(-5)=25\). (b) \(g(f(x))=\frac{x^2-3}{2}\). (c) From \(y=(x-3)/2\) obtain \(x=2y+3\). Hence \(g^{-1}(x)=2x+3\).

(a) \(f(2)=5\), \(h(5)=243\), \(h(1)=3\) and \(f(3)=7\). Difference \(243-7=236\). (b) \(3(2x+1)-2=6x+1\). (c) \(2x+1>3x-2\Rightarrow x<3\). (d) \(3^x=3^{-2}\Rightarrow x=-2\). (e) \(g^{-1}(x)=\frac{x+2}{3}\). (f) \(\frac5{2x+1}+3x-2=\frac{5+(3x-2)(2x+1)}{2x+1}=\frac{6x^2-x+3}{2x+1}\). (g) \(x=f(4)=9\).

(a)(i) \(y=3/(-6)+2=1.5\). (ii) \(y-2=3/x\Rightarrow x=\frac3{y-2}\). (b)(i) \(g(5)=2-5=-3\). (ii) \(h(2)=4\), \(h(4)=16\) and \(h(16)=65536\). (iii) \(2-x=2^3=8\Rightarrow x=-6\). (iv) \(x=g(-1)=3\).

(a) \(g(-2)=4\), then \(f(4)=17\). (b) \(g(f(x))=(5x-3)^2=25x^2-30x+9\). (c) Solve \(y=5x-3\) for x: \(f^{-1}(x)=(x+3)/5\).

(a) \(5-18=-13\). (b) \(5-3(x+2)=-3x-1\). (c) \(5-3(5-3x)=9x-10\). (d) Solve \(y=5-3x\) for x: \(f^{-1}(x)=(5-x)/3\).

(a) \((3x)^2=9x^2\). (b) Solve \(y=3x+5\) for x: \(f^{-1}(x)=(x-5)/3\). (c) \(3(3x+5)+5=9x+20\).

(a) \(f(3)=5\) and \(f(5)=9\). (b) \(g(f(x))=(2x-1)^2+(2x-1)=4x^2-2x\). (c) \(y=2x-1\Rightarrow x=(y+1)/2\), so \(f^{-1}(x)=(x+1)/2\). (d) \(2/x+2/(x+2)=\frac{2(x+2)+2x}{x(x+2)}=\frac{4(x+1)}{x(x+2)}\), where \(x\ne0,-2\).

(a) \(f(2)=8\Rightarrow f(8)=512\). (b) \(g(h(x))=3(2x+1)-5=6x-2\). (c) \(y=2x+1\Rightarrow x=(y-1)/2\), so \(h^{-1}(x)=(x-1)/2\).

(a)(i) \(f(3)=11\). (ii) \(g(3)=8,\ g(8)=2^8=256\). (b) Rearrange y = 2x + 5 and swap variables: \(f^{-1}(x)=(x-5)/2\). (c) \(f(h(x))=2(7-3x)+5=19-6x\). (d) \(1<2x+5\le9\Rightarrow-2<x\le2\). Integers: −1, 0, 1, 2.

(a) \(2^3=8\). (b) \(g(0.5)=2\Rightarrow f(2)=3\). (c) \(f^{-1}(x)=(x+1)/2\). (d) \(2(2x-1)-1=4x-3\). (e) \((2x-1)^2+6=4x^2-4x+7\). (f) x, for x > 0. (g) \(g^{-1}(x)=g(x)\), since taking a reciprocal twice returns x. (h) fh(x), because \(f(h(x))=2(2^x)-1=2^{x+1}-1\).

(a)(i) 5. (ii) \(h(-1)=1/3\Rightarrow f(1/3)=-7/3\). (iii) \(f^{-1}(x)=(x+3)/2\). (iv) \(2(2x-3)-3=4x-9\). (v) For x ≠ −1, multiply by x + 1: \((2x-3)(x+1)=1+2(x+1)\). Expanding and rearranging gives the stated equation. (vi) \(x=(3\pm\sqrt{9+48})/4=(3\pm\sqrt{57})/4\), giving 2.64 or −1.14. (b) \(\frac{(x-1)(x-2)}{(x+5)(x-2)}=\frac{x-1}{x+5}\), with original exclusions x ≠ 2, −5.

(a) \(g(1/2)=1/2\Rightarrow f(1/2)=2\). (b) \(g^{-1}(x)=1-x\). (c) \(h(g(x))=(1-x)^2+1=x^2-2x+2\). (d) \(1-x=7\Rightarrow x=-6\). (e) \(x^2-3x+1=0\Rightarrow x=(3\pm\sqrt5)/2\), giving 0.38 and 2.62. (f) f and g, since f(f(x)) = x and g(g(x)) = x.

(a) \(4-6x\). (b) \(4-3(4-3x)=9x-8\). (c) \(g(-1)=3,\quad g(3)=3^{-3}=1/27\). (d) \(y=4-3x\Rightarrow x=(4-y)/3\), so \(f^{-1}(x)=(4-x)/3\). (e) \(3^{-(4-3x)}=1\) requires \(4-3x=0\), hence x = 4/3.

(a) \(f(2)=4\Rightarrow g(4)=2/5\). (b) \(2=10(x+1)\Rightarrow x=-4/5\). (c) \((6x-2)-[3(x+2)-2]=3x-6\).

(a) \(f(1)=4\Rightarrow h(4)=64\). (b) \(y=(x-1)/4\Rightarrow x=4y+1\), so \(g^{-1}(x)=4x+1\). (c) \((x^3-1)/4\). (d) x = 3.

(a)(i) 8. (ii) \(g(17)=1/2\Rightarrow h(1/2)=2(1/4)+7/2=4\). (b) \(7=(x-3)(x+3)=x^2-9\Rightarrow x=\pm4\). (c) \(2x^2+7x-11=0\Rightarrow x=(-7\pm\sqrt{137})/4\), giving −4.68 and 1.18. (d) \(f^{-1}(x)=(x+2)/5\). (e) \(x=g(-0.5)=7/(-3.5)=-2\).

(a) \(g(5)=1/10,\quad f(1/10)=4.5\). (b) \(g(g(x))=\frac1{2(1/(2x))}=x\). (c) \(y=5x+4\Rightarrow x=(y-4)/5\), so \(f^{-1}(x)=(x-4)/5\). (d) \((1/2)^x=2^{-x}=2^3\Rightarrow x=-3\).

(a) \(g(18)=4,\quad f(4)=4+2/4-3=1.5\). (b) \(y=x/2-5\Rightarrow x=2(y+5)\), giving \(g^{-1}(x)=2x+10\).

(a) \(x=(-1\pm\sqrt{13})/2\approx-2.30,\ 1.30\). (b) \(f(2x+7)=(2x+7)^2+(2x+7)-3=4x^2+30x+53\). Thus p = 4, q = 30, r = 53. (c) \(g^{-1}(x)=(x-7)/2\). (d) \(2^x=2^{-2}\Rightarrow x=-2\). (e) \(h(3)=8,\ h(h(3))=256,\ h(h(h(3)))=2^{256}=1.157920892\ldots\times10^{77}\approx1.158\times10^{77}\).

(a) \(g(6)=36,\ f(36)=75\). (b) \((2x+3)^2=100\Rightarrow2x+3=\pm10\Rightarrow x=3.5\text{ or }-6.5\). (c) \(y=2x+3\Rightarrow x=(y-3)/2\), so \(f^{-1}(x)=(x-3)/2\). (d) \(f^{-1}(5)=1,\quad f(1)=5\).

(a)(i) Read 1.5. (ii) x ≈ 1.2. (iii) The left branch has its maximum between −1 and 0. Thus the largest integer level with 3 crossings is k = −1. (iv) Draw y = x and read its three intersections: x ≈ −2.2, −0.8, 2.3. (b)(i) \(h(3)=8,\ g(8)=1-16=-15\). (ii) \(y=1-2x\Rightarrow x=(1-y)/2\Rightarrow g^{-1}(x)=(1-x)/2\). (iii) \(x^2-1=3\Rightarrow x=\pm2\). (iv) \(1-6x=2x\Rightarrow x=1/8\).

(a)(i) \(9-15=-6\). (ii) \(g(13)=1/2;\ h(1/2)=1/4+5/2=2.75\). (b) \(f^{-1}(x)=(x-3)/4\). (c)(i) \(4x+3=23\Rightarrow x=5\). (ii) \(x^2+5x-7=0\Rightarrow x=(-5\pm\sqrt{25+28})/2=(-5\pm\sqrt{53})/2\approx1.14,\ -6.14\).

(a) \(g(f(x))=2[(x+2)^3-5]+10=2(x+2)^3\). (b) \(y+5=(x+2)^3\Rightarrow x=\sqrt[3]{y+5}-2\Rightarrow f^{-1}(x)=\sqrt[3]{x+5}-2\). (c) \(h(-1/5)=-5;\ g(-5)=0\).

(a)(i) 14. (ii) \(7-2(x-3)=13-2x\). (iii) \((5x)^2-8=25x^2-8\). (b) \(y=7-2x\Rightarrow g^{-1}(x)=(7-x)/2\). (c) \((3x+5)^2-8=9x^2+30x+17\). (d) \(3(3x+5)+5=83\Rightarrow9x+20=83\Rightarrow x=7\). (e) \(6x+10<7-2x\Rightarrow8x<-3\Rightarrow x<-3/8\).

(a)(i) \(f(2)=-3;\ g(-3)=-1/3\). (ii) \((-2)^3+1=-7\). (b) \(1-2/x=(x-2)/x\). (c) \(y-1=x^3\Rightarrow h^{-1}(x)=\sqrt[3]{x-1}\). (d) A, F, D, respectively: decreasing line with positive intercept; reciprocal hyperbola; increasing cubic with y-intercept 1. (e) Apply k to both sides: \(x=k(2)=2^5-3=29\).

(a) \(x=f(2)=4(3)=12\). (b) \(fg(x)=4[(x^3/2-1)+1]=2x^3\). (c) \(y=x^3/2-1\Rightarrow x^3=2(y+1)\Rightarrow g^{-1}(x)=\sqrt[3]{2(x+1)}\).

(a) \(f(-1)=2;\ f(2)=5\). (b) \(gf(3x)=[(3x)^2+1+2]/3=3x^2+1\). (c) \(y=(x+2)/3\Rightarrow x=3y-2;\ g^{-1}(x)=3x-2\).

(a) \(g(3)=11;\ g(11)=43\). (b) \(fg(x)=3(4x-1)+5=12x+2\). (c) \(x=f(11)=3(11)+5=38\).

(a)(i) \(2(6)-3=9\). (ii) \((2x)^3=8x^3\). (b) \((2x-3)^3=125\Rightarrow2x-3=5\Rightarrow x=4\). (c) \(y=2x-3\Rightarrow x=(y+3)/2;\ g^{-1}(x)=(x+3)/2\).

(a) \(2^3=8\). (b) \(2(2x-3)-3=4x-9\). (c) \((2^{x+1})^2=2^{2x+2}=4^{x+1}\).

(a)(i) \(f(2)=7,\quad g(7)=9^2=81\). (ii) \(f(0.5)=2.5,\quad f(2.5)=8.5\). (b) \(y=3x+1\Rightarrow x=(y-1)/3\), so \(f^{-1}(x)=(x-1)/3\). (c) \(f(g(x))=3(x+2)^2+1=3x^2+12x+13\). (d) \(x^2+3x+1=0\Rightarrow x=(-3\pm\sqrt5)/2\approx-2.62,-0.38\).

(a) \(g(1)=-2,\quad f(-2)=1/2\). (b) \(y=x^3+1\Rightarrow x=\sqrt[3]{y-1}\), so \(h^{-1}(x)=\sqrt[3]{x-1}\). (c) \(x^2-3x+2=(x-1)(x-2)=0\), hence 1 or 2.

(a)(i) \(f(2)=6,\ h(6)=39\). (ii) \(4(2/x+1)-2=8/x+2\). (b) \(2/x+1=0.2\Rightarrow2/x=-0.8\Rightarrow x=-2.5\). (c) \(g(3)=5/3,\quad g(5/3)=2/(5/3)+1=11/5=2.2\). (d)(i) \(4x-2=2/x+1\Rightarrow4x^2-2x=2+x\Rightarrow4x^2-3x-2=0\). (ii) \(x=(3\pm\sqrt{9+32})/8=(3\pm\sqrt{41})/8\approx-0.43,1.18\).

(a) \(g(-2)=5,\ h(5)=3^5=243\). (b) \(y=1-2x\Rightarrow x=(1-y)/2\), so \(g^{-1}(x)=(1-x)/2\). (c) \(x=(-1\pm\sqrt5)/2\approx-1.62,0.62\). (d) \((1-2x)^2+(1-2x)-1=4x^2-6x+1\). (e) Apply h to both sides: \(x=h(2)=3^2=9\).

(a) \(g(-1)=4,\ f(4)=27\). (b) \(g(h(x))=(3x+1-1)^2=9x^2\). (c) \(y=(x-1)^3\Rightarrow x=1+\sqrt[3]y\), so \(f^{-1}(x)=1+\sqrt[3]x\).

(a)(i) \(1-2(-5)=11\). (ii) \(3(1-2x)-2=1-6x\). (b) \(x=(5\pm\sqrt{25+44})/2=(5\pm\sqrt{69})/2\approx-1.65,6.65\).

(a) \(f(-2)=6,\quad g(6)=8^2=64\). (b) \(h^{-1}(x)=(x+5)/3\), giving \(h^{-1}(22)=27/3=9\).

(a) \(g(1/2)=2,\ f(2)=-3\). (b) \(gf(x)=1/(2x-7)\), x ≠ 7/2. (c) \(y=2x-7\Rightarrow x=(y+7)/2\), so \(f^{-1}(x)=(x+7)/2\).

(a)(i) 3. (ii) 4. (iii) \(2(2x-1)-1=4x-3\). (iv) \(f^{-1}(x)=(x+1)/2\). (v) \((2x-1)^2=4\Rightarrow2x-1=\pm2\), so x = −1/2 or 3/2. (b)(i) \(y=k/x,\ k=8(2)=16\), so \(y=16/x\). (ii) 32.

(a)(i) 11. (ii) 22. (b) \(y=4x-1\Rightarrow x=(y+1)/4\), so \(g^{-1}(x)=(x+1)/4\). (c) \(6+(4x-1)^2=16x^2-8x+7\). (d) \(4(4x-1)-1=3\Rightarrow16x=8\Rightarrow x=1/2\).