Question

Find the domain and range of the given functions.

1. h(x)= (squareroot4x+1)+3x-5

2. f(x)=3cos(x/3)-1

3. g(x)= log (x+2)

Can someone show me the steps to figure the above problems?

Answer 1

First find the domain of each function. From the domain, we can find what values each function can possible have, this is the range.

We can take advantage of certain rules, or theorems:
The domain of polynomials is ℝ.
The domain of sin, cos is also ℝ.
The domain of √(g(x)) is the domain of g(x) for g(x)≥0.
The domain of ln(g(x)) is the domain of g(x) for g(x)>0.

1. (squareroot4x+1)+3x-5
3x-5 : domain is ℝ
√(4x+1) : domain is the values of x which ensures 4x+1≥0 => x≥-1/4
The domain of the function is therefore the more restrictive of the two, or
&#8477 ∩ [-1/4,∞)
or simply [-1/4,∞)

Similarly, for #3,
3. g(x)= log (x+2)
We look for x for which x+2>0, or
x+2>0 => x>-2.
Therefore dom g(x) = (-2,∞).

I will leave #2 for your practice.

Answer 2

I'm really unsure of how to treat the cos(x/3). YOu stated about that the domain for sin, cos is all real numbers. Would it be (-infinity, infinity) for both domain and range?

Answer 3

Indeed, but for domain only.

For range, you will note from the graphics plot of cos(θ) that it is a wavy curve that varies from -1 to +1, which means that the range of cos(θ) is [-1,1]. The same phenomenon can be said of sin(θ).

Now let's go back to function 2, which is
f(x)=3cos(x/3)-1
the division of (x/3) does not change the range of cos(). However, the multiplier 3 and the subtraction of -1 does.

Can you find the range of f(x), knowing the range of cos(x/3) is [-1,1]?

Answer 4

It changes the graph. So the range is expanded more. Would the new range be [-4,2]?

Answer 5

Sorry MathMate,
The anonymous answer above was mine. Just on a different computer.

Answer 6

Correct!

Answer 7

Thank you so much for your help!!

Answer 8

composite funtions