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Studying for a test? Prepare with these 13 lessons on Derivative rules.

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

Video transcript

In terms of k, where
k does not equal 0, what is the y-intercept of
the line tangent to the curve f of x is equal to 1/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. And the graph f of
x is equal to 1/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 would look
something like that. And then on the negative side,
it looks 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/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 1/k. We can visualize the line
tangent to the curve there. So it 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 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. If we could figure out the
slope of the tangent line, we already know that
line contains the point. Let me do this in
a different color. We know it contains
the point k comma 1/k. So if we know its slope, we
know what point it contains, we can figure out what
its y-intercept is. So the first step is
just, 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 writing it as 1/x, I'll write it 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 of k is going
to be equal to negative k to the negative 2 power. Or another way of thinking about
it, this is equal to negative 1 over k squared. So this right over
here is the slope of the tangent
line at that point. Now, 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 plus b. 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/k when x is equal to k. So this is going to be
equal to negative 1-- that's not the same color. Negative 1 over k
squared times k plus b. Now, what does this simplify to? See, k over k squared is
the same thing as 1/k, so this is going
to be negative 1/k. So this part, all of this
simplifies to negative 1/k. So how do we solve for b? Well, we could just
add 1/k to both sides and we are left with-- if
you add 1/k here-- actually, let me just do that. Plus 1/k, left-hand side,
you're left with 2/k, and on the right-hand side,
you're just left with b, is equal to b. So we're done. The y-intercept of
the line tangent to the curve when x equals
k is going to be 2/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/k squared x plus our
y-intercept, plus 2/k. And we're done.