Calculus Ap exam Questions 1980?

Question

Hey I am in Calculus right now and my teacher gave us this packet with questions from previous Ap exams.

1980 AB3

Let f(x)=ln(x^2) for x>0 and g(x)=e^(2x) for x>(or equal to)0. Let H be the composition of f with g, that is, H(x) = f(g(x)), and let K be the composition of g with f, that is K(x)=g(f(x)).

(a) Find the domain of H and write an expression for H(x) that does not contain the exponential function.

(b) Find the domain of K and write an expression for K(x) that does not contain the exponential function.

(c) Find an expression for f^-1(x), where f^-1 denotes the inverse function of f, and find the domain of f^-1.

I would really appreciate your help. I need answers by tonight, thanks.

-Also I need to show work so if you can show how you got it I would appreciate it, but if you know the answer, or just one of the answers, I will accept that aswell, thanks.

Answer 1

a) H(x)=ln((e^2x)^2)
bring the 2 at the end of the equation to the front
H(x)=2ln(e^2x)
the natural log and e cancel out
H(x)=2(2x)
H(x)=4x

D:(0,+infinity)

b)K(x)=e^(2ln(x^2))
put the 2 in front of the natural log behind the x^2
K(x)=e^ln(x^2)^2
the e and natural log cancel out
K(x)=(x^2)^2
K(x)+x^4

D:(0,+infinity)

c)f(x)=ln(x^2)
y=ln(x^2)
x=ln(y^2)
e^x=e^(ln(y^2))
e^x=y^2
square root of e^x=y

the domain of ln(x^2) is x>0, which makes the range all real numbers. when you inverse an equation, the domain and range get flipped around. that is, the domain of the inverse of the equation (square root of e^x=y) is all real numbers and the range is x>0. but who cares about the range of that equation?

D:all real numbers

Answer 2

(a)
Domain: (0,∞)
H(x) = ln((e^2x)^2) = ln(e^4x) = 4x

(b)
Domain: (0,∞)
K(x) = e^(2ln(x^2)) = e^ln(x^2) * e^ln(x^2) = x^2 * x^2 = x^4

(c)
y = ln(x^2)

x = ln(y^2)

e^x = y^2
___
√e^x = y

I'm not positive about the Domain... I believe it is either (0,∞) due to the restrictions for f(x) or (-∞,∞) because all real numbers provide a value for y.

Answer 3

a) The zeros occur whenever the numerator equals zero. x^3 - 3x^2 - 4x + 12 = 0 Factor by grouping. x^2 (x-3) - 4(x-3) = 0 (x-3)(x^2 -4) = 0 (x-3)(x-2)(x+2) = 0 x = 3, 2, or -2 b) I don't see what p has to do with the problem. H(x) = f(x)/x-3 Should there be a p somewhere in H(x)? c) Cannot do without part b.

Answer 4

(a)

H(x) = f(g(x)) = ln((e^(2x))^2) = ln(e^(4x)) = 4x

The domain is x>0.

(b)

K(x) = g(f(x)) = e^(2ln(x^2)) = e^(4ln(x)) = x^4

The domain is x>0.
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