SHSAT Algebra – Function Problems

Let`s go over some problems concerning functions so the first one what are the roots of f of X equals x squared minus 81 so find the roots for that function the next one would be what is the domain of H of T equals as the square root of 3 minus 70 and number three which regions make up the domain of H of x equals x over the square root of x squared minus nine so find out which of these regions 1 2 3 and 4 make up the domain it could be more than one alright so why don`t you guys take a minute to do each one so the first one you can pause and try it out.

What are the roots of this function so f of x equals x squared minus 81 when they say roots remember what the quadratic roots were it was the x value or x-values that we would plug in to get 0 so in this case we want the x values that makes f of X equal to 0 so all we need to do is set x squared minus 81 equal to 0 well this is a difference of perfect squares so we can factor it as X minus 9 times X plus 9 equals 0 which gives us the roots x equals 9 and x equals negative 9 these two roots give you f of x equals 0 so those two would be f of 9 equals F of negative 9 equals 0 so those are the answers 9 and negative 9 all right so why don`t you try number 2 pause and come back all right what is the domain of H of T equals the square root of 3 minus 70 so the domain would be all the T values that are possible well we have a square root sign so we know that no negative sign can be under so I propose that we find out where 3 minus 7 he is less than zero so less than zero not less than or equal to because it can be equal to so just less than zero let`s try that three minus 70 less than zero so we will treat it like a any quality so we can subtract three subtract three so we get that says negative 70 is less than negative three and now we divide by negative seven to get T is greater than three over seven remember we divided by negative so we flip the sign so these values of T would make three minus 17 negative so the domain would be everything that`s opposite of that so T is less than or equal to three over seven and that`s our domain because those values are possible and we can check it out so let`s try and see what values here what makes sense so we make t equal to five and then we should get a negative under the square root sign let`s test it out so three minus seven times five so that would be the square root of three minus 35 which is the square root of negative 33 and so that`s a negative number under the square root sign so we know that that doesn`t work and now here we can find number less than three over seven so let`s try zero if we plug in zero for that T we get the square root of three so it works so the domain is T is less than or equal to three over seven and it`s important to note that you can write the domain in different ways so that`s one way of writing it you can also write it as well if it`s less than three or equal to three over seven we can say it`s between negative infinity and three over seven this bracket here means equal to 3 over 7 so it`s between negative infinity and three over seven including three over seven.

Great and the reason why you don`t put brackets over or right next to negative infinity or infinity is because nothing can ever be equal to that so always put parentheses all right so now let`s try the third problem so which regions make up the domain of H of x equals x over the square square root of x squared my it`s nine so take some time to try that and then come back alright now that we`re back let`s check it out so we`re trying to find what the domain is so we can try out points here but let`s just solve it or use the function to find the domain so we have H of X equals two x over the square root of x squared minus nine so now this square root sign cannot have a negative number underneath it but also it`s in the denominator so cannot equal zero so that we know our domain must have this following condition x squared minus nine must be greater than zero and once we solve for this we know our domain another way of doing it is just finding what region cannot be in the domain and then taking the opposite of that or whatever the leftover regions are but let`s do it this way so we add nine on both sides and we get x squared is greater than nine and take the square root of both sides and so now we have X is equal to three and negative three when you take the square root of a number you can get the square root as a positive and negative number that`s important to know so that means that if x squared equals sixteen then x equals 4 and negative 4 because 4 squared is 16 and negative 4 squared is 16 so here are our two roots or our two values our two endpoints so if we plug in any of these numbers then we get zero for this so then what`s our domain since we have two two roots here 3 and negative 3 we need to try out each one separately or you can just use inspection to see that X must be greater than 3 and less than negative 3 it`s not the case that X is just greater than both of them because since we got the negative 3 we have to flip signs but let`s actually use the number line to figure this out if we use a number line we can just plug in certain points in each region and see which one`s work well since we have 3 and negative 3 we put those points here that`s where they go and if it were 2 and negative 2 then we would put those points there because that`s where the function is 0 so let`s try region 1 let`s try the number negative 4 so if we plugged in negative 4 for H of negative 4 what would we get let`s find out we would get negative 4 over radical 16 minus 9 and that would be an OK number that would be 4 over the square root of 7 so that`s a positive number and now we know that that region works out so we can shade it in and it`s greater than negative 3 so it would be an open circle and now we can try region – we can try negative 1 for example so let`s try negative 1 since it`s that in that region and so that would be negative 1 over square root of 1 minus 9 and that is negative 1 over negative 8 so that does not work so region 2 does not work so no let`s try region 3 we can try in the number 1 so let`s do H of 1 and we would get positive 1 over 1 minus 9 which is the same thing so that does not work and so we know that region 3 is invalid that does not work and now we try region 4 and we could try the value 4 so H of 4 what is that that would be 4 over square root of 16 minus 9 which is what we got in the beginning which is 4 over the square root of 7 and that works so the regions that work are the two end regions so these two with an open circle so less than negative 3 and greater than 3 and the way to write that would be like this X is greater than 3 and X is less than negative 3 alright or you could use that notation I showed before from negative infinity to negative 3 exclusive and then from 3 to infinity ok great so we went over those problems and now we can move on to the next topic after this thanks for watching the videos please be sure to check out our book on Amazon too special cycle test prep book this book comes with five full-length exams with answer explanations along with a manual to go ahead and better prepare you for the exam you will find the link on the video or in the description above have fun studying.

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