Question

1. Solve the equation x = (2x+ 3)1/2

2. Consider f1(x) = ln(x + 1) + ln (x-1) and f2(x) = ln(x^2-1).

a) State domains and ranges of f1 and f2.

b) Sketch the curves y = f1(x) and y = f2(x).

Answer 1

1.
`x=3`; 2. Domain:
`x>1` Range: all real numbers

Explanation:

Let's find the solutions.

1. Solve the equation
`x=√(2x+3)` so:

`(x)^2=(√(2x+3))^2`

`x^2=2x+3`

`x^2-2x-3=0`

`x1=\frac{-b+\sqrt{b^(2)-4ac}}{2a}`

`x1=\frac{2+\sqrt{(-2)^(2)-(4*1*(-3))}}{2*1}`

`x1=3`

`x2=\frac{-b-\sqrt{b^(2)-4ac}}{2a}`

`x2=\frac{2-\sqrt{(-2)^(2)-(4*1*(-3))}}{2*1}`

`x2=-1`

Although we have two answers, remember that from the original equation the result of
`√(2x+3) > 0` is never negative. So -1 do not solve the equation.

In conlcusion, the equation is solved by x=3.

2A. Domains and ranges of f1(x) and f2(x)

`f1(x)=ln(x+1)+ln(x-1)`

Using logarithmic property
`ln(a)+ln(b)=ln(a*b)` we have:

`f1(x)=ln(x^2-1)` because:

`ln(x)` is defined by
`x>0` then:

`x^2-1>0`

`x>√(1)` so the domain of f1(x) is
`x>1`

Now for the range:

`f1(x)=ln(x+1)+ln(x-1)`

`y=ln(x^2-1)`

`e^y=x^2-1`

`√(e^y+1)=x^2-1` notice that
`e^y+1` is always positive, so the range of f1(x) is all real numbers.

Be aware that although point number two of the problem mentioned two equations, f1(x)=f2(x) by logarithmic properties, so their domains and ranges are the same.

2B. Graph of f1(x) is attached. Because f1(x)=f2(x) both functions plot equal.