AP BC calculus question?

Question

i have a slight problem solving this because i cant understand part a

give the function f defined by f(x) = ln(x^2-9)

a) describe the symmetry of the graph of f (how do i do this?)

b) find the domain of f (i think it is x>-3 and x>3 but how do i write this together?)

c) solve values ln(x^2-9) = 0 (i dont know what to do with ln)

d) write formula for f^-1x (this one i just dont get)

Answer 1

a) describe the symmetry of the graph of f (how do i do this?)

f(x) is symmetric about the vertical line x = 0.  You can find this by graphing the function or by noticing that the function is almost ln(x).  If you know what the shape of ln(x) is that is a good start, but instead of an x inside the ln you have x² - 9.  Since the domain is x > 3 and x < -3, this means that the function will be shaped like ln(x) from (3, ∞) with a vertical asymptote at x = 3 instead of x = 0.  Similarly, the function will be shaped like ln(x) headed to -∞ instead of ∞ from (-∞, -3) with a vertical asymptote at x = -3.  Take a look at the graph to see this clearly if you are uncertain.

b) find the domain of f

The domain of f(x) is x < -3 and x > 3.  Then the domain is (-∞, -3) U (3, ∞).

c) solve values ln(x² - 9) = 0

Remember that ln is log base e.  Also recall that a^b = c is the same as log_a(c) = b where _ denotes the base.  So, in your case

ln(x² - 9) = 0

log_e(x² - 9) = 0

e^0 = x² - 9

1 = x² - 9

1 + 9 = x²

10 = x²

±√10 = x

So x = √10 and x = -√10 are values that solve the equation.

d) write formula for f^(-1)(x)

This is called the inverse function.  To find the inverse function you switch x and y and solve for y.  

y = ln(x² - 9)   ...now switch x and y

x = ln(y² - 9)   ...solve for y

x = log_e(y² - 9)

e^x = y² - 9

e^x + 9 = y²

±√(e^x + 9) = y

So, f^(-1)(x) = ±√(e^x + 9)

Hope this helps you!