### Video Transcript

Discuss the continuity and
differentiability of the function π at π₯ is equal to zero given π of π₯ is equal
to negative nine π₯ minus six if π₯ is less than zero and π of π₯ is π₯ squared
minus nine π₯ minus six if π₯ is greater than or equal to zero.

Weβre given a piecewise function
defined as π of π₯ is negative nine π₯ minus six if π₯ is less than zero and π of
π₯ is π₯ squared minus nine π₯ minus six if π₯ is greater than or equal to zero. Letβs first sketch the functions π¦
is negative nine π₯ minus six and π¦ is π₯ squared minus nine π₯ minus six on the
same graph to get an idea of what weβre looking at.

As we can see, the functions π¦ is
π₯ squared minus nine π₯ minus six and π¦ is negative nine π₯ minus six both meet at
the point negative six on the π¦-axis. So that our lower function π of π₯
is defined differently either side of π₯ is equal to zero. As π₯ approaches zero, both
functions tend to the same value.

Weβre asked to discuss the
continuity and differentiability of our function π of π₯ at π₯ is equal to
zero. So letβs define what we mean by
continuity and differentiability and go from there.

A function π of π₯ is continuous
at π₯ is equal to π (1) π of π exists, (2) the limit as π₯ approaches π of π of
π₯ exists, and (3) the limit as π₯ tends to π of π of π₯ is equal to π of π. A function π of π₯ is
differentiable at π₯ is equal to π if π prime at π₯ is equal to π exists. For a function to be differentiable
at a point π₯ is equal to π, a function must be continuous at that point, although
the reverse is not necessarily the case.

Weβre asked to discuss the
continuity and differentiability of the function at π₯ is equal to zero. So letβs first look at the
continuity at π₯ is equal to zero. Does our function satisfy the three
criteria for continuity?

At π₯ is equal to zero, our
function is defined as π₯ squared minus nine π₯ minus six. So at π₯ is equal to zero, π of π₯
is π of zero, which is zero squared minus nine times zero minus six. And thatβs equal to negative
six. So our first condition for
continuity is satisfied. π of zero exists, and itβs equal
to negative six.

Now letβs check our second
condition for continuity. Does the limit as π₯ tends to zero
of π of π₯ exist? We know that the definition of our
function differs either side of π₯ is equal to zero. So weβre going to need to consider
the behavior of π of π₯ either side of zero and look at both the left-hand and the
right-hand limits.

The left-hand limit, which is the
limit as π₯ tends to zero from the negative direction, is the limit as π₯ tends to
zero from the negative direction of negative nine π₯ minus six, which is negative
nine times zero minus six. And thatβs negative six.

The right-hand limit is the limit
as π₯ tends to zero from the positive side, which is the limit as π₯ tends to zero
from the positive direction of π₯ squared minus nine π₯ minus six. And thatβs equal to zero squared
minus nine times zero minus six, which is negative six.

Since both the left-hand and the
right-hand limits exist and are equal β in fact, the function approaches negative
six from both sides β our second condition for continuity at π₯ is equal to zero is
satisfied. And since the limit as π₯ tends to
zero of π of π₯ is equal to π of zero, which is negative six, our third condition
for continuity is also satisfied. We can therefore say that π of π₯
is continuous at π₯ is equal to zero.

So now letβs discuss the
differentiability of our function at π₯ is equal to zero. Remember that a function π of π₯
is differentiable at π₯ is equal to π if π prime of π exists. Remember that π prime of π₯ is dπ
by dπ₯ in Leibniz notation, which is d by dπ₯ of negative nine π₯ minus six for π₯
less than zero and d by dπ₯ of π₯ squared minus nine π₯ minus six for π₯ greater
than or equal to zero.

Using the fact that the derivative
of a constant times π₯ is the constant and the derivative of a constant is zero, d
by dπ₯ of negative nine π₯ minus six is negative nine. To differentiate π₯ squared, we can
use the power rule. This tells us that the derivative
with respect to π₯ of ππ₯ to the power π is equal to πππ₯ to the π minus
one. In other words, we multiply by the
exponent and subtract one from the exponent.

So the derivative of π₯ squared
with respect to π₯ is two π₯. And we already know that the
derivative of negative nine π₯ is negative nine. So the derivative of π₯ squared
minus nine π₯ is two π₯ minus nine. The derivative of our left-hand
function is negative nine, a constant, since the function is a straight line. So as π₯ approaches zero from the
left, the derivative is negative nine.

π prime at π₯ equal to zero is two
times zero minus nine. Thatβs π₯ is equal to zero
substituted into our right-hand function. And again, thatβs negative
nine. So π prime of zero does exist and
it equals negative nine. So π prime of π₯ exists. Therefore, our function π of π₯ is
both continuous and differentiable at π₯ is equal to zero.