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# Worked example: Tangent to the graph of 1/x

## Video transcript

in terms of K where K does not equal zero what is the y-intercept of the line tangent to the curve f of X is equal to 1 over X at the point of the curve where X is equal to K so let's just think about what they're asking so if I were to draw myself let me draw some quick axes right over here so that's my y-axis this is my x-axis right over here x axis and the graph f of X is equal to 1 over X would look something like this so it looks something like this so it kind of spikes up there and then it comes down and then it goes like this and I'm just doing a rough approximation of it so it looks something like that and the negative side it looks something like something like this so this is my hand-drawn version of roughly what this graph looks like so this right over here is f of X is equal to 1 over X now we are concerning ourselves with the point x equals K so let's say I mean it could be anything that's non zero but let's just say that this is K right over here so that is the point K K 1 over K a 1 over K and we want to find the line we can think about this we can visualize the line tangent to the curve there so it might look something might look something like this might look something like this and we need to figure out its y-intercept where does it intercept the y-axis so we need to figure out this point this point right over here well the best way to do it if we can figure out the slope of the tangent line here the slope of the tangent line is just the derivative of the line at that point and we could figure out the slope of the tangent line we already know that that line contains the point we already know it contains the point this point right over let me do this in a different color we know it contains the point K comma 1 over K so if we know its slope we know what point it contains we can figure out we can figure out what its y-intercept is so the first step is this well what's the slope of the tangent line well to figure out the slope of the tangent line let's take the derivative so if we write f of X instead of writing is 1 over X I'll write as X to the negative 1 power that makes it a little bit more obvious that we're about to use the power rule here so the derivative of F at any point X is going to be equal to well it's going to be the exponent here is negative 1 so negative 1 times X to the now we decrement the exponent to the negative 2 power or I could say it's negative x to the negative 2 now what we care about is the slope when x equals K so f prime F prime of K is going to be equal to is going to be equal to negative K to the negative 2 power to the negative 2 power or another way of thinking about it this is equal to negative 1 over K squared negative 1 over K squared so this right over here is the slope of the tangent line at that point now let's think about let's just think about what the equation of the tangent line is and we could think about it in slope intercept form so we know the equation of a line in slope intercept form is y is equal to MX plus b where m is the slope and b is the y-intercept so if we can get it in this form then we know our answer we know what the y-intercept is going to be it's going to be B so let's think about it a little bit this equation so we could say Y is equal to our M our slope of the tangent line when X is equal to K we just figure out to be this business it equals this thing right over here so let me write that in blue negative 1 over K squared times X times X plus V so how do we solve for B well we know what Y is when X is equal to K and so we can use that to solve for B we know that Y is equal to 1 over K Y is equal to 1 over K when X is equal to K so this is going to be equal to I'm going to keep the colors consistent equal to negative 1 that's not the same color negative negative 1 over K squared times K times K plus B plus B then white plus B now what does this simplify to see k over k squared is the same thing as 1 over K so this is going to be negative 1 over K so this part all of this all of this simplifies to negative 1 over K so how do we solve for B well we could just add 1 over K to both sides we could just add 1 over K to both sides and we are left with if you add 1 over K here actually let me just do that plus 1 over k plus 1 over K left hand side you're left with 2 over K 2 over K and on the right hand side you're just left with B is equal to B so we're done the y intercept the y-intercept of the line tangent to the curve when x equals K is going to be is going to be 2 over K if we wanted the equation of the line well we've done all the work let's write it out it'll be satisfying it's going to be Y is equal to negative 1 over K squared X X plus plus our y-intercept plus 2 over K + 2 / plus 2 over K and we're done