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Interactive video lesson plan for: relations and functions vertical line test domain range ordered pairs

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Good day students welcome to mathgotserved.com in this clip were going to be going over relationship functions forget to visit our website@mathgotserved.com for access to a wide variety of math tutorials ranging from algebra all the way to calculus to get started let's take a look at what on the definition of relation is so is go ahead and write that down relation so what are relation and mathematics relation are basically set of order to pairs okay so a set of order pairs is a relation okay so for that take a look at an example of what the order here is so let's go ahead and take a look at this we have one to two three and one five now this is an example of the relation because the have a set of ordered pairs okay now what you notice one point out want to note is a special relationship between of function in relation okay so before we write down the relationship let's define what the function is first okay so let's define that functions what is the definition of a function you function is the order here okay special time to repair so a function is in relation what relation is relation are susceptible repairs so function is a relation or you can see function in a set up here back@every in you can reverted input out the X or the domain okay so our final every inputs of X or any element in the domain exactly why all students okay now this can be viewed at the why or the elements of the rate right so you want to ensure that every you in this assigned exactly what so this can be assigned multiple inputs that's find the area input is cited that we one outputs that's what the definition of the function is okay so the points to know that this is the side know is that all functions are relation okay but not but not all relations our functions okay so what you going to be doing X is the going to be looking at different representations of relation in the going to be relation a strategy to determine if they are functions are not so let's go ahead and take a look at the first representation which is ofRight students is go ahead and take a look at Tom the remaining review problems on the chain rule so for question number four we are to find F prime of X is F of X is equal to sign square two X minus five divided by X square times the natural log of them of X so we can notice here that we have quotient of two functions let's call the numerator function you and the denominator function the to how to find the derivative of the quotient of two functions so that the have you over be right what will the derivative the so is simply going to use the quotient rule okay so the quotient rule it's we know that is the you prime minus you UV prime divided by square okay so that's the quotient rule right so all you have to do here is fine you prime in v prime and then substitute it into this formula and that will give us the result that we desire okay let's start with you to you is equal to sign square of two X minus five now this is a composite function is a composition of three functions okay we have the square function the sign derivative function and in the linear function X minus five so we can rights you as F of G of H of X okay F of G of each of X so this is F of G of H of X and you prime applying the chain rule to the F prime of G of H of X times G prime of H of X times each prime of X okay so let's go ahead and of the compose you Indian we can find the derivative of the different functions in plug it into the general formula okay so use F of G of H of X which is equal to all sine square of two X minus five right taking how the F so F of X is X square then G of X actually G of HMX is going to be sine of two X minus five we can see how this function can't result a result of the composition of these two okay with ethology G of X is sine X and on the limits is of H of X the innermost function is all two X minus five okay now let's go ahead and find the derivative of these three functions F prime of X is to X find the power rule you prime of X using trigonometric differentiation is cosine X and then H prime of X is simply to okay now what we going to do is obliges resulting Got to all the chain rule formula right so U prime is going to be at prime of G of H of X is together that simply take G roots G I'm sorry G of H of X limit take that okay here okay and then to get this to that in the U prime is going to be all to the X is getting two times sine of two X minus five right to that's F prime of G of H of X times G prime of H of X is simply take H and plug it into G prime of X so that yields is the cosine X will have cosine two X minus five times H prime of X which is to our right side is the oldest cases into this chain rule formula to generate this expression right here okay have one piece of the chain rule formula the U prime now let's go ahead and of fi

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