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# Solving equations graphically (1 of 2)

CCSS.Math:

## Video transcript

graphs of a of x equals e to the X and R of X is equal to 1 over X times X minus 1 times X minus 2 are shown below estimate the solution to e to the X so that's G of X being equal to essentially R of X within 0.01 so we want to figure out we want to figure out for what value does G of X equal R of X equal R of X and they want us to estimate it we can either just try to get as close as we can from this graph they want us to be within 0.01 and we can also use a calculator or kind of try numbers out to hopefully zero in on this point right over here where a of X is equal to R of X so what I want to do is let me draw a little table here let's try out some X values and then for each of these X values let's see where we land on a of X on a of X and where we land on R of X and where we land on R of X and then we can decide whether we are too high or too low and I encourage you to pause this video before I actually go ahead and do this and try to do this on your own but I do suggest using some form of calculator or probably a calculator well I'm assuming you've given a go at it now I will attempt it now just eyeballing it and eyeballing it is helpful because that'll give us kind of our first order approximation of what at what x value are these two functions equal if we look if I just look at this graph the way it's drawn it looks like this is pretty close to two point one pretty close to two point and why it looks like when X is two point one both of those functions look pretty close to I don't know this looks like about seven point seven point seven or seven point eight or something like that but let's let's figure out what they're doing so let's see when X is equal to two point one and get my calculator out when X is equal to two point one e of X is just well e of X is just e to the X power so e to the two point one power is equal to eight point one six six so let me write that down eight point one six six and what is our of X our of X is 1 divided by X so that's going to be two point one times X minus one well that's going to be one point one so that's times one point one times X minus two well that's just going to be 0.1 times point one and that is equal to four did I do that right now that can't be it's two point one it's two point one over two point one times one one point one times point one one over all of that so do four point three to see two point one R of X is four point I guess that's possible actually that looks right because R of X declines so sharply right over here so it's actually at two point one where actually R of X is actually closer to right over right over right over here give or take so it's equal to four point three two so a two point one y of X is actually a much larger value than R of X so e of X is clearly a of X is clearly too high R of X has already dropped a good bit by then if I were to go all the way down to two at two it looks like R of X it looks like actually R of X kind of spikes up it's a it's it just it just it just goes to infinity as we approach to so we're not going to go all the way down to two but why don't we lower this a little bit why don't we try to point zero five so at two point zero five what is e of X e to the e of X is e to the X right so e to the two point zero five power it gets us seven point seven six I'll round it eight seven point seven six eight approximately seven actually all of these are approximate so I'll just write seven point seven six eight and what is R of X I'll just keep rounding to the thousands here we didn't - well we didn't have to round to much just because that was so far off but I'll put it there it was three - actually it was three to nine so I could let me write it this way it's three to nine zero so let me throw that nine here just so everything we evaluate the function to the thousands so let's evaluate R of X when we're at two point zero five it's going to be 1 divided by X which is now two point zero five times X minus one which is one point zero five times X minus two which is 0.05 and that gets us to nine point two nine I'll round to to nine point two nine two so nine point two nine two so now we're on this side where R of X is roughly right over here and it's more than a of X which is at seven point seven which is right around here so now we've our X values too low so let's see maybe let's see if we can go a little bit higher and we'll go let's try to go roughly halfway between these two and but I don't want to get too precise because you have to get to the nearest hundredth so let's go to two point zero seven two point zero seven so e to the two point zero seven two point zero seven is equal to seven point nine two five if I round it seven point nine two five seven point nine two five I want to do all of this in green just to be color consistent and now let's evaluate R of X at that same value so one divided by X which is two point zero seven times that minus one which is one point zero seven times that minus two which is zero point zero seven which gives us six point four four I guess we could say six point four five zero six six point four five zero so at two point oh five that was too low two point oh seven is too high now R of X has dropped a of X so two point so we know the right answer is in between these two numbers and so if we select two point zero six that's definitely going to be within 0.01 of the right answer so I would go I would go with two point two point zero six is going is definitely going to be but within 0.01 of the correct solution to this but just for fun let's actually just try it out so e to the two point zero six is seven point eight four or I guess we could round to six and if we were evaluate R of X its 1/2 point zero six times that minus one which is one point zero six times zero point zero six gets us to seven point six three two so we're also getting pretty pretty close but our precision that they gave they don't say that these have to equal they have to be within each other of that they say let's estimate the solution so there's some actual precise solution to this right over here some x value that where these are actually equal to each other that's that's the x value which gives us this point of intersection we just have to get within two 0.01 of that x value and two point zero six definitely does the trick