Video Transcript
Simplify the function π of π₯ equals π₯ plus one times π₯ plus two over π₯ plus three times π₯ plus four times π₯ plus two, and determine its domain.
It actually makes more sense to begin by determining the domain of the function first. Remember, the domain of a function is the set of possible inputs to that function that yield real outputs. And of course when dealing with the domain of a rational function, we need to be really careful to ensure that the denominator of that function is never equal to zero.
So we can determine the domain of the function by first finding the values of π₯ that do make the denominator zero. In other words, what values of π₯ satisfy the equation π₯ plus three times π₯ plus four times π₯ plus two equals zero? Well, since this is the product of three expressions, we know that, for the product to be zero, either one of the expressions must themselves be equal to zero. That is, π₯ plus three equals zero or π₯ plus four equals zero or π₯ plus two equals zero. To make π₯ the subject in each case, weβre going to subtract the constant. In our first equation, we subtract three, and we find π₯ is negative three. In our second, we subtract four, so π₯ is negative four. And in our third, we subtract two from both sides. So we get π₯ equals negative two.
Otherwise, the function is the quotient of two polynomials. And the domain of any polynomial function is the set of real numbers. So the domain of our function is the set of real numbers not including the set of π₯-values that make the denominator zero. Thatβs negative two, negative three, and negative four. But what is the function π of π₯ when we simplify it?
Well, to simplify a function, we look for common factors in the numerator and denominator. We see there is in fact just one common factor, and that is π₯ plus two. So we can divide through by π₯ plus two, leaving π₯ plus one over π₯ plus three times π₯ plus four. We might choose to leave it like this, or we can distribute the parentheses on the denominator. To distribute the parentheses, we multiply the first term in each expression. π₯ times π₯ is π₯ squared. We then multiply the outer terms to get four π₯ and the inner terms to get three π₯, giving us a total of seven π₯. Finally, we multiply the three by four, and that gives us 12. So we see that the expression for π of π₯ is equivalent to π₯ plus one over π₯ squared plus seven π₯ plus 12. And so we have our value for π of π₯ and the domain of the function.
Now, we might be wondering why we chose to find the domain of the function before we simplified the expression. Suppose we had simplified the fraction first. We would have ended up with the expression π₯ plus one over π₯ plus three times π₯ plus four. In this case, we know that π₯ cannot be equal to negative three, nor can it be equal to negative four. But we might have deduced that π₯ can be all real numbers other than these. And we know that if π₯ is equal to negative two, weβre doing zero divided by zero, which is undefined. And so we must work out the domain of our function before we simplify it any further.
π of π₯ is π₯ plus one over π₯ squared plus seven π₯ plus 12. And the domain is the set of real numbers not including the values of π₯ in the set negative two, negative three, or negative four.