BAM!!! Mr. Tarrou. Well, we're continuing that
work of getting ready for the AP
calculus exam. And in this short video
we're going to be taking derivatives of
absolute value functions. In the
description you will find a link to the
full review playlist. All these questions
inspired by multiple choice questions
i've used in the past four years. And
you'll also find links to any
full-length lessons that I've done that
apply to these concepts. Now I have done
a lesson on finding derivatives of absolute
value functions. In that lesson, I
worked on rewriting these equations in
terms of... sort of like... Well, when you take
an expression and you square it and then
square root it, that acts the same as an
absolute value function. And we approach
those problems that way. But
this time we're going to look at it a
little differently. We're going to
rewrite these equations into a piecewise
function, and then take the derivative of
piecewise function. You know couple years
of experience of teaching calculus is
learned that's a little bit easier, or
maybe a lot easier, than how I did my
first lesson. So we're going to walk through the
first example. And because this is a
review, will give you some time to
practice number eight on your own. Then
just reveal solution a little bit at a
time. So the first thing you want to do
when you find the derivative of these
absolute value functions, if you want to
go through the process of this piecewise
function idea instead of a
square root of a squared value, is... and
it's you know true regardless of what
technique you use... You have to remember
that these absolute value functions are
all going to have a sharp bend at the
values generally that makes the
expression inside the absolute value
equal to 0. Kinda like a critical value.
I'm going to take this 2x minus 3,
we're going to set it equal to 0. And of
course solve it. That's going to be x is
equal to 3/2. And that's the
break. And again that's just a simple linear
function. This absolute value function is going to have
a sharpened at the x value of 3/2,
or 1.5. So we're going to say then
that y is equal to...
well let's see here. If we put a number
in that is larger than 1.5 like 2, we're
going to have 2 times 2 is equal to 4...
minus 3 which is equal to 1. And the
absolute value 1 is just 1. So the
absolute values don't really seem to be
doing anything. It tells us how far we are
away from zero on the number line. But
really numerically nothing happens.
So we just go okay... so we have 2x minus
three plus one. And that's going to be
our function for all X's that are.... well...
going to be... And I kind of... I want
my conditions can go from low to high, so
actually the absolute value symbols did
nothing per say, when X was greater than
or equal to 3/2's. If you do put in 3/2,
then 2 times 3/2 is equal to
three... and then 3 minus 3 is equal to 0.
And the absolute value zeros is also
just zero. There's a sign change. But when
you plug in an x value which is less
than 3/2, let's just pick 0 for
convenience, 2 times 0 is 0. The absolute value of
negative 3, there is a sign change, and the
actual answer of course is equal to
three. So we show the effects of that
absolute value function if we are rewriting
this in pieces, as introducing a
parentheses with a multiplication
negative 1. Just like we would have done
solving absolute value equations back in
algebra 1 or algebra 2. Whenever it was
you were exposed to it first. So absolute
value... changing the sign when x is less
than three halves. Of course we can
distribute this negative combine, like
terms and we have y is equal to.... It is now a
piecewise function... negative 2x.
Negative times negative is positive, and then 3
plus 1 is equal to four. And that's when
x is less than 3/2. And we have
2x-2 for when x is greater than or
equal to 3/2. And now we just have a
nice little simple linear function in 2
pieces to take the derivative of. So y
prime is equal to negative 2 when x is
less than three halves. And the
derivative of 2x minus 2, which is equal
to 2... when x is greater than 3/2.
Do you see what I just did there? Not
only as I do this I think man in the
previous lesson I would've wrote this as
y is equal to the square root of 2x
minus 3 squared plus 1, and then go to
that derivative process. And it was a lot more
complicated than it is by rewriting it
in terms of a piecewise function.
Secondly notice that I'm saying that the
domain of the original function, because
the domain of this absolute value
function is all real numbers. So i'm
showing that in the piecewise function,
but when i get down to the derivative
I've lost the equal sign of course.
Because y prime is not, the derivative
for this function, is not equal to 2 at
three halves. Of course because again,
that's a sharp bend. You can't find the
derivative at a sharp bend. Take this
idea and see if you can't find the
derivative for this absolute value
function. A little more complicated because
it's not a linear function. And i'm also
going to reveal the solution one step at
a time. Every few seconds reveal the next
step, in case you get stuck and you want to
pause it, and then finish it up. And we'll
also look at what the graph of the
derivative looks like as well.
Coming up, now try it first, right now!
Now first of all love my chalkboard a
graph paper grid there. And I always
get all kinds of people ask me well how
do I draw these dots! If this is a
parabola, now in my Algebra one class i'm
talking about graphing right now of
parabolas with transformations. And we
have a basic parent function shifted to
the right 2 and down 4, so we have a
parabola opening up. But of course the
absolute value, this is all inside the
absolute value symbol or function, so
anytime this expression comes out to be
negative of course the absolute value is going to make
a positive. And our graph looks something
like that.
Now I can do maybe a t-table get a
slightly better accurate picture of how
sharp the slope is there to the left and
right of 0 and 4. But now comes the
graphing of piecewise functions. All
these things we learned. It so cool. It all keeps
coming back in and getting used
again. We have just a y intercept of
negative 4 and a slope of 2. So 1 2 3
4, & up 2 over 1. Now
ok, this is only defined when x is less
than zero. So even though you know we
have a line, like course like, this and
I'm speaking for anybody watching this
that is not necessarily taking calculus.
Because this is a review of graphing
piecewise functions. So we have a
function, a line with slope of 2 and y intercept of -4.
Going to be true all the way up until when
x is less than zero, and then between... x being
in between 0 & 4...
Did you notice that in that middle part
of this piecewise function I included
the negative sign. And how I know of
course the absolute value is going to
take care of those negatives in and
switch the sign of our answer comes in
two ways, either I can you know plug-in
an x value of say
1. Right, because we have those two
critical points along the x-axis where
the sharp bends are for the absolute
value function. So I know something's
happening to the left of zero, between 0
and 4, and to the right. And i can plug in
points. Like I can say, well let's let X
be 1, and 1-2 is such-and-such, and work
out the arithmetic. Or just understand of
course we are just talking about a
parabola and we're in calculus now. We should
know what that thing looks, like knowing
that graph was falling below the x-axis
their requiring us to introduce that negative
set of parentheses to change the sign
like the absolute value symbol would do. So
between 0 and 4 then, we have this
line y equals basically negative 2x plus
4. So now our y-intercept is positive 4.
but of course we can take a derivative
of the sharpen so that's undefined. Slope
is negative 2, so down 2 over 1, down
2 over 1, and just keep on going.
Making a nice solid line until we get to
that x value of four where we have that
other sharp bend. And then picking right
back up, when x is greater than 4 we
didn't have that sign change, so we are back to
that 2x-4 line. So coming away back over
here, up 2 over 1, up 2 over 1, up 2 over 1.
Up 2 over 1, pick up with that open dot.
And voila, a little review of dealing
with piecewise functions as well as
parabolas using... graphing parabolas
using transformations. I'm Mr. Tarrou.
Let's go get that 5, like BAM!