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

Answer:

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

Step-by-step explanation:

Let's find the solutions.

1. Solve the equation $x=\sqrt{2x+3}$ so:

$(x)^2=(\sqrt{2x+3})^2$

$x^2=2x+3$

$x^2-2x-3=0$

$x1=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$

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

$x1=3$

$x2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

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

$x2=-1$

Although we have two answers, remember that from the original equation the result of $\sqrt{2x+3} > 0$ 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:

$x^2-1>0$

$x>\sqrt{1}$ so the domain of f1(x) is $x>1$

Now for the range:

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

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

$e^y=x^2-1$

$\sqrt{e^y+1}=x^2-1$ 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.