Explain how to determine which numbers must be excluded from the domain of a rational expression. Please provide an example with
1 answer:
Answer:
- When we are having a rational expression i.e. a expression of the type:
Where f(x) and g(x) are polynomial functions.
Now the domain of this rational expression is whole of the real numbers except the points where the function g(x) will be zero.
Hence we have to exclude the points where the given denominator quantity is zero.
- Let us consider an example as:
Let R(x) denote the rational function as:
Now the domain of this rational function will be whole of the real line minus the points where the denominator is zero.
We know that (x-2)(x-3) is zero when x=2 or x=3.
Hence, the domain of R(x) is: R- {2,3}.
You might be interested in
Answer:
Step-by-step explanation:
No. of days of study = 6
Amount of food to use for the 6 days = 72 mm
Given that equal amount of food was used each day, amount used per day =
This would enable us derive a function for the amount of food remaining f(x), which can be expressed as a function of the number of days that have passed, (x).
Thus, we would have:
Let's check if this equation expresses the relation of the function well.
At the end of day 1, if we used, 12mm, we would have 72 - 12 = 60mm of food remaining.
If we replace x in the equation we derived, we would get the same 60mm as the amount of food remaining.
If you try 2, 3, 4, 5 days, you'd get same thing.
At the end of the study, after the 6th day, the food would remain nothing.
Replace 6 for x in the relation function derived, you'd get a value of 0.
Lets check with every option
(a) [-4,-3]
We plug in -4 for x and -3 for x
f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.
(b) [-3,-2]
We plug in -3 for x and -2 for x
f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.
(c) [-2,-1]
We plug in -2 for x and -1 for x
f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.
(d) [-1,0]
We plug in -1 for x and 0 for x
f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.
(e) [0,1]
We plug in 0 for x and 1 for x
f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.
(f) [1,2]
We plug in 1 for x and 2 for x
f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.
so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]
Answer:
Step-by-step explanation:
So we have the two functions:
And we want to find f(g(1)).
So, let's find g(1) first:
Substitute 1 for x:
Simplify:
Add:
So:
Now, substitute 2 for x in f(x):
Multiply:
Add:
So: