Discrete Mathematics with Applications 4th Edition Chapter 7 - Functions - Exercise Set 7.1 - Page 394 14

Answer

$ J_{5} = \left \{ 0, 1, 2, 3, 4 \right \}, h :J_{5}\rightarrow J_{5} \,\,and\,\, k :J_{5}\rightarrow J_{5}\\ h (x)=(x + 3)^{3}\,\,mod\,\,5\,\,and\,\,k(x)=(x^3+4x^{2} + 2x + 2)\,\,mod\,\,5.\\ h (0)=(0 + 3)^{3}\,\,mod\,\,5=27\,mod\,5=2\\ h (1)=(1 + 3)^{3}\,\,mod\,\,5=64\,mod\,5=4\\ h (2)=(2 + 3)^{3}\,\,mod\,\,5=125\,mod\,5=0\\ h (3)=(3 + 3)^{3}\,\,mod\,\,5=216\,mod\,5=1\\ h (4)=(4 + 3)^{3}\,\,mod\,\,5=343\,mod\,5=3\\ k(0)=(0^3+4(0)^{2} + 2(0) + 2)\,\,mod\,\,5=2mod\,5=2\\ k(1)=(1^3+4(1)^{2} + 2(1) + 2)\,\,mod\,\,5=9mod\,5=4\\ k(2)=(2^3+4(2)^{2} + 2(2) + 2)\,\,mod\,\,5=30mod\,5=0\\ k(3)=(3^3+4(3)^{2} + 2(3) + 2)\,\,mod\,\,5=71mod\,5=1\\ k(4)=(4^3+4(4)^{2} + 2(4) + 2)\,\,mod\,\,5=138mod\,5=3\\ so\,\,\\ h(0)=k(0)=2 \\ h(1)=k(1)=4 \\ h(2)=k(2)=0 \\ h(3)=k(3)=1 \\ h(4)=k(4)=3 \\ \therefore h=k $

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