 Continuing our discussion about algebraic functions in lecture three here, I wanna point out some differences about solving and evaluating algebraic functions compared to what we did previously with like geometric and numerical representations of functions. For an algebraic function, evaluations, it's generally a simple task of arithmetic. We saw this in the previous video. If ever you need to evaluate the function at some value x, you find the corresponding y coordinate just by plugging x into the function and you simplify it. And so this makes, for example, finding y coordinates, y intercepts, specifically super easy, to find the y intercepts, you just have to evaluate f of zero, which again becomes a task of arithmetic, usually worst case scenario right there. Solving an equation is a little bit more challenging, right? So evaluation, you're given the x and you wanna find the y. But when it comes, in the other situation, when you are given the y and you wanna find the x, you have to solve an equation. You have to solve for x, the equation f of x equals zero. This can be a little bit more challenging. And therefore, as an example, finding x intercepts, you have to solve the equation f of x equals zero. That in general is a much more challenging question. And the difficulty of solving this equation f of x equals zero or f of x equals zero or f of x equals y, honestly the two are not really any more difficult from than each other. The difficulty of solving that equation really depends on the nature of the function itself. Linear equations are gonna be pretty easy to solve, quadratic and polynomial go a little bit harder, rational ones, radical ones, logarithmic ones, exponential ones, these are all function families we'll talk about this semester. And so the nature of solving it depends on the function itself. So we're gonna start off because we're at the beginning of our course here with a very, very simple example. Take the function f of x equals two x plus five. This is what we call a linear function. We'll talk about some more about linear functions in the not too distant future. So let's show how one evaluates and one solves a function that's linear, right? Linear evaluation is pretty nice. If you wanna find f of negative three, you're just gonna replace x with negative three in your formula. And then arithmetically simplify this thing. Two times negative three would give us a negative six. Plus five to that would give us a negative one. And so that is the function evaluation. Not so bad. Now if we want to solve set equation, right? So solve f of x equals negative three. So what we're trying to figure out right now is what x-coordinate will produce the y-coordinate negative three. So in that case, if we take f of x is equal to negative three, we unravel what f of x here means. It means two x plus five equals negative three. So then we start using laws of algebra to help us solve this thing. So for example, if we minus negative, or if we minus five from both sides of the equation, we get that five minus five will cancel out. And then on the right-hand side, we get negative three minus five, which will give us a negative eight. Therefore, two x equals negative eight. And then to get rid of the two that's attached to the x will divide both sides by two. Whatever we do to the left-hand side, we have to do to the right-hand side as well. I like to often say the phrase, what's good for the goose is good for the gander. In order for the equation to be balanced, we have to do the same thing to both sides. Now, why are we dividing by two? Well, because division by two is the inverse operation to multiplying by two. And so that'll help us solve the equation just by undoing one operation on the left-hand side step by step by step. We then end up with x equals negative four over two which simplifies to be x equals negative four. And so this would be the solution to the equation for which you can check set solution by plugging it back into the equation. Notice, if you evaluate at negative four, the function, you'll get two times negative four plus five, which is gonna be negative eight plus five. Negative eight plus five is equal to negative three. This verifies that we have the correct solution. And so when it comes to evaluating and solving for functions in this context and algebraic function, that's what one does, to evaluate you plug in the x and simplify. When it comes to finding x given the y, you're gonna have to solve some algebraic equation which will depend on the function itself. This linear equation we did because it's fairly simple, but as this course progresses, our equations might get more challenging, more challenging, more challenging, but never too challenging for us because we'll take incremental steps before we just dump into the deep end of the pool there.