Respuesta :

SOLUTION:

Case: Piecewise functions

Method:

a) The graph:

The equations:

Domain:

[tex](-\infty,4)\cup[4,\infty)[/tex]

b) x-intercepts

[tex]\begin{gathered} x^3-16x=0 \\ x(x^2-16)=0 \\ x(x-4)(x+4)=0 \\ x=0,x=4,x=-4 \\ But\text{ }x<4 \\ Hence \\ x=0,x=-4 \end{gathered}[/tex]

Also

[tex]\begin{gathered} -log_4(x-3)+2=0 \\ log_4(x-3)=2 \\ x-3=4^2 \\ x-3=16 \\ x=16+3 \\ x=19 \end{gathered}[/tex]

The x-intercepts are:

x= 0, x = 19, x= -4

c) The intervals where g(x) is positive include:

[tex][-4,0]\cup[4,19][/tex]

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