Question
1] Find the domain of the following functions_f(x) =V-x2+3x + 10 [15 pts] 4 b) f(x) =Er"-I-1 [15 pts] f(r)-75 ~u-Tr [15 pts]
1] Find the domain of the following functions_ f(x) =V-x2+3x + 10 [15 pts] 4 b) f(x) =Er"-I-1 [15 pts] f(r)-75 ~u-Tr [15 pts]
Answers
Find the domain of the function f given by each of the following. $$ f(x)=\frac{1+x}{3 x-15 x^{2}} $$
We need to find the domain in here. There are no restrictions. No matter what X value we plug in, we're going to get a Y value. So our domain consists of all numbers, from negative infinity to positive infinity.
Or go for this problem is to find the domain of each function. So first we have one over one plus you. To the acts Well notice that the domain here can only be restricted by the denominator. If this was undefined and the only way this could be undefined, this fraction would be if the denominator equal zero, Meaning each of the X has to equal and negative one. In order for this to equal zero. And we know this, there's no way this can equal zero. So therefore the domain is all real numbers. There's never a point where it's on the fine. Now, if we look at 1/1 minus E to the X story because we see that um if X equals zero, then we have one minus one. So now the denominator is undefined or the denominator zero. And the functions undefined. So where do we go from negative infinity to zero And then from 0 to Infinity, but not including zero.
Let's say we had the equation y two equals X minus for over. Expert was two x minus 15. Since we're dividing by a variable, you know we can't divide by zero, so you do not want X squared plus two x minus 15 to be equal to zero. You don't want that. This is fact herbal. It's a quadratic well, multiplies the negative 15 and asked a positive to positive five in negative three so we wouldn't want the Factor X Plus five to be equal to zero. Nor would we want the factor x my mystery to be equal to zero. So if he's attracted five and add three, that's telling us, Try it again, that we wouldn't want extra Pete. You're negative five or positive three. Because if it was, we wouldn't be divided by zero. You can't do that. So the domain, everything else is going to be fair game. We can put anything from negative infinity up to negative five not included from negative 5003 included. Not included. Excuse me and from three deposited Anthony included. Those are all be eligible inputs into the domain of this function
In this problem, I want to find the domain for the function F. Of X equals the quantity five X plus three. Over the quantity four X minus one. So the first part here is my numerator, the five X plus three. And that has a domain of all real numbers. And then the second part is the denominator, the four X -1. And there is a restriction on this. We cannot have a denominator that's equal to zero. So I need to figure out what value of X would make it equal to zero. So I'm gonna solve this out. So X Is going to equal 1/4. So any number, Any real number can be used in this spot for ex except for 1/4 because 1/4 makes it zero and that's not allowed. So we can have X can be less than 1/4. X can be greater than 1/4 but X cannot equal 1/4. So an interval notation, the domain looks like negative infinity to 1/4, parentheses. E and parentheses E. 1/4 to infinity. And these are parentheses because the 1/4 is not included in the domain.