#(a)
The given function is
[tex]f(x)=3-\sqrt[]{10-2x}[/tex]
Since there is no square root for a negative value, then put (10 - 2x) greater than or equal zero to find all values of x can make the function f(x) defined
[tex]10-2x\ge0[/tex]
Add 2x to both sides
[tex]\begin{gathered} 10-2x+2x=0+2x \\ 10\ge2x \\ 2x\leq10 \end{gathered}[/tex]
Divide both sides by 2
[tex]\begin{gathered} \frac{2x}{2}\leq\frac{10}{2} \\ x\leq5 \end{gathered}[/tex]
The domain is all values of x less than or equal to 5
[tex]D=(-\infty,5\rbrack[/tex]
#b)
The given function is
[tex]g(x)=\frac{17x-1}{x^2-4x+3}[/tex]
Equate the denominator by 0, then factorize it into 2 factors to find the values of x which make the denominator equal to 0, then exclude these values from the real number
[tex]\begin{gathered} x^2-4x+3=0 \\ (x-3)(x-1)=0 \end{gathered}[/tex]
Equate each factor by 0
[tex]\begin{gathered} x-3=0 \\ x-3+3=0+3 \\ x=3 \end{gathered}[/tex][tex]\begin{gathered} x-1=0 \\ x-1+1=0+1 \\ x=1 \end{gathered}[/tex]
The domain of g(x) is all real numbers except 1 and 3
[tex]D=(-\infty,1)\cup(1,3)\cup(3,\infty)[/tex]
