If 𝑓 and 𝑔 are two real functions, where 𝑓 𝑥 is equal to 𝑥 minus one over 𝑥 squared plus three 𝑥 minus four and 𝑔 𝑥 is equal to 𝑥 plus three, determine the values of 𝑓 plus 𝑔 minus four, if possible.
You can try and solve this problem. Well first we need to know what the question is actually asking. So if we first look at this part here, it says 𝑓 plus 𝑔. What that means is if we add together our two functions, and then this negative four means if we substitute negative four for 𝑥, the same we can say if we add together our two functions and we substitute in negative four for 𝑥, what is our value gonna be, if it’s possible?
When working out this kind of question, to work this kind of problem out, we can either add together our functions before we substitute in negative four or we can actually do that after we’ve substituted in negative four. For this question, I’m just gonna actually work out each function with 𝑥 equal to negative four first and then add them together afterwards.
So as I said, the first thing we’re gonna do is substitute in 𝑥 equal to negative four; that gives us negative four plus three, so we can say that our function 𝑔𝑥 will be equal to negative one when 𝑥 is equal to negative four. Okay right, we can have a look at the other function now, 𝑓𝑥.
Okay and again we have 𝑓𝑥; we’re gonna substitute in 𝑥 is equal to negative four. So on the numerator, we have negative four minus one, and on our denominator, we have negative four all squared plus three multiplied by negative four and minus four. Okay now we can simplify, which will give us negative five as the numerator and then 16 plus negative 12 minus four on the denominator, which gives us the final answer of negative five over zero.
Right, having zero as the denominator means that this number is actually undefined, and it means that if we add our two functions together, we’ll still have negative one plus an undefined number, which therefore means that 𝑓 plus 𝑔, so our function of 𝑓 plus function of 𝑔, when 𝑥 is equal to negative four, is undefined.