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Consider f which is defined
for all real numbers. At what arguments x is f
of x not differentiable? So to think about
that, I'm actually going to try to visualize what
f prime of x must look like. So I'm going to do f prime
of x in this purple color. So if we look at f
of x right over here, it looks like its slope is
pretty much consistently negative 2 over this interval
between x equals, I guess, it's like negative 8
and 1/2 all the way up to x equals negative 2. It looks like the slope
is a constant negative 2. So if I were to
draw its derivative, its derivative would
look something like this. Its derivative looks
something like this. But then something interesting
happens at x equals negative 2. Right as we cross x
equals negative 2, it looks like the
slope goes from being negative to being positive. And it looks like
right out the get go, if I were to estimate the
slope of its tangent line, it starts changing. It's not a line anymore. It's a curve. The slope of the tangent
line right at this point looks like it's
around-- I don't know-- it looks like it's
around 3 and 1/2. Because if I were to draw a
tangent line right over here, it looks like if I move
1 in the x direction, I move up about 3 and
1/2 in the y direction. So I'm just trying to,
obviously, estimate it. So it looks like the
slope goes up to 3 and 1/2 right when I cross that point. And then the slope becomes
lower and lower and lower all the way until I get to
this point right over here, all the way until I
get to x equals 2. And it looks like it continues
to get lower all the way until you get to x equals 3. So it looks like the
slope of the line is-- it looks like it's getting
lower at a constant rate, I guess I could say. So it looks like
it's doing something like this over this interval. But then right as x crosses
3, this becomes a flat line. The slope is 0 here. So right as x crosses
3, the slope becomes 0. So we immediately
see there are points where it looks like
the slope jumps. And at these points
we really don't have a defined derivative. The slope jumps there as well. And so at what arguments
is f not differentiable? Well, it's not differentiable
when x is equal to negative 2. When x is equal to
negative 2, we really don't have a slope there. Remember, when
we're trying to find the slope of the
tangent line, we take the limit of the
slope of the secant line between that point and some
other point on the curve. If we did that as we
approached from the left, it looks like the
derivative is negative 2. If we do that from the right,
it looks like the derivative is something like
positive 3 and 1/2. And so we're not getting the
same limit of the secant line as we approach from the left and
as we approach from the right. And the same thing is
happening at x is equal to 3. At x equals 3, as we
approach from the left, the slope looks like
it is decreasing. It is approaching-- I don't
know-- maybe around negative 1. But as we approach
from the right it looks like the slope is 0. So we do not have the same
limit of the secant slope as we approach from the
left- and right-hand sides. So at both of these points
we see the derivative jump, and it looks like f of
x is not differentiable.