This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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1(a) f(x)=sqrt(x) Domain: Square root requires non-negative argument. x 0 Domain: [0,) Range: Step 1: Let y=f(x). y=sqrt(x) Step 2: Square both sides. y^2=x Step 3: y 0 (principal square root). Step 4: x=y^2 0, which is in domain for all y 0. As y goes from 0 to , x goes from 0 to . Range: [0,) [0,) 1(b) g(x)=2^x Domain: Exponential defined for all real x. D=(-,) Range: Step 1: Let y=g(x). y=2^x Step 2: y>0 always since 2^x>0. Step 3: x=_2 y, defined for y>0. As x -, y 0^+; as x , y . Range: (0,) (0,) 2(a) m(x)=x^2 Domain: R (polynomial). Range: Step 1: Let y=m(x). y=x^2 Step 2: x=± sqrt(y). Step 3: Real x requires y 0. For all y 0, real x exists. Minimum y=0 at x=0; y as |x|. Range: [0,) [0,) 2(b) n(x)=x^3 Domain: R. Range: Step 1: Let y=n(x). y=x^3 Step 2: x=[3]y. Step 3: Cube root defined for all real y. Bijective from R to R. Range: (-,) (-,) 3. u(x)=sqrt(x^2-4) Domain: Inside square root 0. x^2-4 0 x^2 4 x -2 or x 2 Range: Step 1: Let y=u(x). y=sqrt(x^2-4) y 0 Step 2: Square both sides. y^2=x^2-4 x^2=y^2+4 Step 3: |x|=sqrt(y^2+4) 2, since y^2+4 4. Always true for y 0. Minimum y=0 at |x|=2; y as |x|. Range: [0,) [0,) 4. v(x)=(x^2-1) Domain: Argument >0. x^2-1>0 |x|>1 Range: Step 1: Let y=v(x). y=(x^2-1) Step 2: x^2-1=e^y x^2=e^y+1>1 (since e^y>0) Step 3: Always true for all real y. As |x| 1^+, x^2-1 0^+, y -. As |x|, y. Range: (-,) (-,)