Benjamin W. answered 04/12/23
Chemistry PhD student, Lab Technician
All of these problems boil down to domain and range. Domain is the acceptable input of a function and the range is the set of outputs. Domain is the one to watch out for because if there is a fraction bar or a square root we can get undefined outputs. E.G. 1/x where x is zero and √-x where x is negative.
To solve problem three we can eliminate D) as false because √x is undefined at negative numbers making the graph look different. C) can also be eliminated because both functions are defined from negative infinity to infinity. A and b are both true and the answers, just make sure to test b to make sure it shrinks by 1/2. a is true because you can input any number and there is no undefined area in either function.
for problem 5 we can eliminate k(x) because using y = mx+b the slope is -21 and just subtracting y values we can also eliminate h(x) with a slope of .25 and f(x) because the value decreases at certain points. G(x) is an exponential function and this can be proved by x3=125 taking respective cube roots we get 5 as a base and plugging in we see it works.
To solve problem six, we just check to see if there are any square roots or inputs in the denominator. There are none in any of these functions so the range is defined everywhere. A graphing utility confirms this as the final answer that none of the functions have a range of (-2 to infinity).
I hope this helps!