Graph natural log of x square root of x^2-1

Problem

y=ln(x√(,x2−1))

Solution

1. Identify the domain of the function. For the natural logarithm to be defined, the argument must be positive: x√(,x2−1)>0 Since the square root is non-negative, we require x>0 and x2−1>0 This results in the domain x>1

2. Simplify the expression using logarithm properties to make it easier to analyze. Use the product rule ln(a*b)=ln(a)+ln(b) and the power rule ln(an)=n*ln(a)

y=ln(x)+ln((x2−1)(1/2))

y=ln(x)+1/2*ln(x2−1)

3. Determine the vertical asymptote. As x approaches 1 from the right, x2−1 approaches 0 so ln(x2−1) approaches −∞ There is a vertical asymptote at x=1

4. Find the derivative to determine the slope and behavior.

d(y)/d(x)=1/x+1/2⋅(2*x)/(x2−1)

d(y)/d(x)=1/x+x/(x2−1)

d(y)/d(x)=(x2−1+x2)/(x*(x2−1))

d(y)/d(x)=(2*x2−1)/(x*(x2−1))

5. Analyze the derivative for x>1 Since 2*x2−1>0 and x*(x2−1)>0 for all x in the domain, the derivative is always positive. The function is strictly increasing.

6. Evaluate the end behavior. As x→∞ both ln(x) and ln(x2−1) approach ∞ so y→∞

7. Locate the x-intercept by setting y=0

x√(,x2−1)=1

x2*(x2−1)=1

x4−x2−1=0

Using the quadratic formula for x2

x2=(1+√(,5))/2

x=√(,(1+√(,5))/2)≈1.272

Final Answer

y=ln(x√(,x2−1))

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