1. For this question f is the function f(x) = 3 + 2/x . Which of these statements about f is True?

Question

a. f has an inverse function, it's the function with defining equation g(x) = 2/(x-3)
b. f does not have an inverse function
c. f has an inverse function, it's the function with defining equation g(x) = 2/x - 3
d. f has an inverse function, but we don't know how to find the defining equation of the inverse function explicitly

2. For this question f(x) = x2 + 1 . Which of these statements about f is True?
a. f has an inverse function, it's the function whose defining equation is g(x) = sqrt(x-1)
b. f has an inverse function, but we don't know how to find the defining equation of the inverse function explicitly.
c. f doesn't have an inverse function
d. f has an inverse function, it's the function whose defining equation is g(x) = sqrt(x+1)

3. For this question f(x) = x3 . Which of these statements about f is True?
a. f has an inverse function, it's the function g(x) = cube root of (x)
b. f doesn't have an inverse function
c. f has an inverse function, it's the function g(x) = x/3
d. f has an inverse function, but we don't know how to find the defining equation of the inverse function explicitly

4. For this question, f(x) = (3+x)/2. Which of these statements is True about f?
a. f has an inverse function, it's the function g(x) = 2x-3
b. f doesn't have an inverse function
c. f has an inverse function, it's the function g(x) = x/2 - 3
d. f has an inverse function, but we don't know how to find the defining equation of the inverse function explicitly

Answer 1

The simple rule of inverse functions is that when one function is substituted as the argument of its inverse, then the result will be the sole variable, X.



So, the easiest way to find a simple inverse function is to let Y=F(x), and then solve for X:



f(x) = 3 + 2/x

y = 3 + 2/x

y-3 = 2/x

x = 2/(y-3) <<---- this is your inverse function!



g(x) = 2/(x-3) This corresponds to answer (A) for the first question.



All of the subsequent questions can be solved in the same manner.

Answer 2

Write g(x,e)=f(x+e)-f(x). it is non-quit under the product topology relative to the subset {(x,e) in R^2;a<=x<=b and a<=x+e<=b}, so the set A={(x,e);f(x+e)>f(x)} ={(x,e);0

Answer 3

1) A

2) A

3) A

4) A