Topic 1 - Final

Question 1

1.1.

Find the domain and range of the following function

y = (x^2 + 9)/(x-3)

Answer 1

Factor and you’re going to cancel out the two x -3 and end up with

x+3

So there is a hole at x = 3 and you plug in this value into f(x) to find the y value, and the range is all real numbers BUT this y value

Question 2

1.1. Domain and range of composite functions

Answer 2

In order to find the domain for a function like f x g, both f and g and the new function you create have to have the same domain

In order to find the domain for a function like f(g(x)), g(x) has to be defined and this new composite function but not necessairly f(x)

Question 3

1.1 What is the domain of

sqrt (1/x + 1)

Answer 3

You set 1/x + 1 ≥ 0

1/x ≥ -1
now you have to consider for when x is negative and when it is positive

for when postiive (x>0)
x≥-1
so this inequality holds true for all positive x values

for when negative (x<0)
x≤−1 which means that for any negative value, the root is only holding true when x≤−1

By consequence the domain must be (-inf, -1] U (0, inf)

Question 4

1.4.
Vertical asymptote of tan function

Answer 4

x = pi/2 + kpi

Question 5

1.4.

What is the caveat of restricting teh domain to make a function 1 to 1 to take the inverse

Answer 5

The restricted domain still has to include the ENTIRE range of the original function

Question 6

Find value of constant k such that the domain of

f(x) = 1/(sqrtkx^2-x) is (-inf, 0) U (5, inf)

Answer 6

You know that the denominator cannot equal 0 and it also cannot be less than 1

so factor it out
x(kx -1) and the roots are 0 and 1/k

If k is negative then the parabola opens downwards and the values where it is positive are between teh roots 0 and 1/k, which doesn’t make sense.

So k has to be positive, and in order to fit the constraints of the domain where it is undefined, that would be where

1/k = 5 which is when k = 1/5

Question 7

1.4

Inverse of trig functions

Answer 7

One thing to note is that the domains are the ranges of all the functions

Arcsin

Domain [-1, 1]
Range [-pi/2, pi/2]

Arccos

Domain [-1, 1]
Range [0, pi]

Arctan
Domain (-inf, inf)
Range (-pi/2, pi/2)

Question 8

1.5

Find the domain of
ln(ln(ln(x)))

Answer 8

For the inside function ln(x)
x>0

For the 2nd function ln(ln(x))
ln(x)>0
bring both sides up by e and you’ll get
x>1

For the 3rd function ln(ln(ln(x)))
lnln(x))>0
same thing bring both sides up by e
ln(x)>1
repeat
x>e

Question 9

1.1

Domain of

sqrt(1/x + 1)

Answer 9

This is a bit of a tricky one but you have to split it up into two cases and you end up with

(-inf, -1]U(o, inf)

one important thing to note is that when you take the case assuming x<0, it just flips the sign, yuo are not multiplying -x by -1

Question 10

How do you find the inverse of

3f(2x+4) -1 using a table of values

g^-1 (2)

Answer 10

So you want to set the expression equal to 2 and rearranging you get

f(2x + 4) = 1

this expression holds true when x is equal to 2 and since (2x+4) represents the f(x) then you know that

2x+4 = 2
so x = -1