Explicación paso a paso:
5) f(4) si f(x) = [tex]\sqrt{x^{2}-4 }[/tex]
[tex]f(4) =[/tex] [tex]\sqrt{(4)^{2}-4 }[/tex]
[tex]f(4) = \sqrt{16-4}[/tex]
[tex]f(4) = \sqrt{12}[/tex]
[tex]f(4) = 2\sqrt{3}[/tex]
6) g(-2) si g(x) = [tex]\frac{4x-1}{x+2}[/tex]
[tex]g(-2) = \frac{4(-2)-1}{-2+2}[/tex]
g(-2) = se encuentra diviendo por 0 asi que es indefinida
7) h(3) si h(x) = [tex]2-\frac{4x}{3}[/tex]
[tex]h(3) = 2-\frac{4(3)}{3}[/tex]
[tex]h(3) = 2-\frac{12}{3}[/tex]
[tex]h(3) = 2-4[/tex]
[tex]h(3) = -2[/tex]
8). p(-1/2) si p(x) = [tex]\frac{2}{x^{2} -1}[/tex]
[tex]p(-\frac{1}{2}) = \frac{2}{(\frac{1}{2})^{2}-1 }[/tex]
[tex]p(-\frac{1}{2}) = \frac{2}{\frac{1}{4}-1 }[/tex]
[tex]p(-\frac{1}{2}) = \frac{2}{\frac{1-4}{4} }[/tex]
[tex]p(-\frac{1}{2}) = \frac{2}{\frac{-3}{4} }[/tex]
[tex]p(-\frac{1}{2}) = -\frac{8}{3}[/tex]
9) m(5) si m(x) = [tex]\frac{3x+2}{\sqrt{4-x} }[/tex]
[tex]m(5) = \frac{3(5)+2}{\sqrt{4-5}}[/tex]
[tex]m(5) = \frac{15+2}{\sqrt{-1} }[/tex]
[tex]m(5) = \frac{17}{\sqrt{-1}}[/tex]
m(5) = se encuentra indefinida
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