graph this function and determine the domain,range and vertical asymptote.
f(x)=In(4-x) no need to graph just assist with points
Here is a nice little page that will give you a quick graph
Just type
log(4-x) into the first window, and click on "Draw"
http://rechneronline.de/function-graphs/
Remember, we can only take log of positive numbers
so 4-x > 0
-x > -4
x < 4
So there is a vertical asymtpote at x = 4
the domain is x < 4
range: y ∊ R
to get the graph, on your calculator find some critical points, e.g.
x = 3.999 -- (3.999 , -6.9)
x = 0 ---- (0, 1.39)
x = 3 --- (3 , 0)
x = -5 --- (-5 , 2.2)
x = -1000 --- (-1000 , 6.9) ---> just to show what happens at the far left.
To determine the domain of the function f(x) = ln(4-x), we need to consider the restrictions on the natural logarithm function.
The natural logarithm function ln(x) is defined only for positive values of x. Therefore, in our case, the expression 4-x inside the ln() must be positive.
Setting 4-x > 0 and solving for x, we get:
4 - x > 0
x < 4
Hence, the domain of f(x) = ln(4-x) is all real numbers x such that x < 4.
To determine the range of the function, we need to consider the behavior of the natural logarithm function. The range of ln(x) is all real numbers, which means the range of f(x) = ln(4-x) will also be all real numbers.
Finally, let's determine the vertical asymptote. The vertical asymptote is a vertical line that the graph approaches but doesn't touch. Since the natural logarithm function ln(x) is not defined for x ≤ 0, the vertical asymptote of f(x) = ln(4-x) occurs when the expression 4-x equals zero.
Solving 4-x = 0 for x, we get:
x = 4
Therefore, the vertical asymptote of f(x) = ln(4-x) is x = 4.
Although I can't directly graph the function here, you can plot some points to get an idea of its shape. Choose different values for x such that x < 4, substitute them into the function f(x) = ln(4-x), and calculate the corresponding y-values using a calculator or a computer program. Repeat this process for a few values of x. For example:
- For x = 3: f(3) = ln(4-3) = ln(1) = 0
- For x = 2: f(2) = ln(4-2) = ln(2)
- For x = 1: f(1) = ln(4-1) = ln(3)
This will help you get an idea of how the graph behaves.