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?
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
ℝ ∩ [-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.
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?
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]?
It changes the graph. So the range is expanded more. Would the new range be [-4,2]?
Sorry MathMate,
The anonymous answer above was mine. Just on a different computer.
Correct!
Thank you so much for your help!!
composite funtions
Certainly! Let's go through each function one by one and determine the domain and range.
1. h(x) = √(4x+1) + 3x - 5
To find the domain of h(x), we need to determine the values of x for which the function is defined. The square root function (√) is defined for non-negative real numbers. Therefore, the expression inside the square root (4x+1) must be greater than or equal to zero:
4x + 1 ≥ 0
To solve for x, we subtract 1 from both sides and then divide by 4:
4x ≥ -1
x ≥ -1/4
So the domain of h(x) is all real numbers greater than or equal to -1/4.
To find the range of h(x), we need to consider the possible output values. Since h(x) includes a square root and a linear term (3x - 5), the range will vary depending on the value of x. However, since the √(4x+1) term is always non-negative, the minimum value of h(x) will occur when the linear term (3x - 5) is minimized. This happens when x is at its minimum value of -1/4, which we found to be the lower bound of the domain.
Substituting x = -1/4 into the function:
h(-1/4) = √(4(-1/4)+1) + 3(-1/4) - 5
= √0 + (-3/4) - 5
= -3/4 - 5
= -20/4 - 5
= -25/4
Therefore, the range of h(x) is all real numbers greater than or equal to -25/4.
2. f(x) = 3cos(x/3) - 1
The domain of f(x) is all real numbers since the cosine function is defined for any input.
The range of f(x) is typically between -1 and 1, since the range of the cosine function is [-1, 1]. However, f(x) is shifted downwards by 1 unit, so the range of f(x) is [-2, 0].
3. g(x) = log(x+2)
To find the domain of g(x), we need to consider the argument of the logarithm function, which is (x+2). For the logarithm function to be defined, the argument must be greater than zero:
x + 2 > 0
To solve for x, we subtract 2 from both sides:
x > -2
So the domain of g(x) is all real numbers greater than -2.
The range of g(x) is all real numbers since the logarithm function can take any positive real value as its input.
I hope this explanation helps you understand how to determine the domain and range of these functions!