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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).

1 Answer

4 votes

Answer:

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.

1. Solve the equation x = (2x+ 3)1/2 2. Consider f1(x) = ln(x + 1) + ln (x-1) and-example-1
User CuriousYogurt
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