### Video Transcript

Let π of π₯ be the
piecewise-defined function which is equal to five π plus ππ₯ squared if π₯ is less
than negative two. π of π₯ is equal to five when π₯
is equal to negative two. And π of π₯ is equal to ππ₯ minus
three π when π₯ is greater than negative two. Determine the values of π and π
so that π is continuous at π₯ is equal to negative two. What can be said of the
differentiability of π at this point?

The question is asking us to first
determine the values of π and π which make the function π continuous at the point
π₯ is equal to negative two. The question then wants us to check
if the function π will be differentiable at the point π₯ is equal the negative
two. We can start by asking what makes a
function continuous at a point.

We recall that a function π is
continuous at a point π if the limit of the function as π₯ approaches π can be
evaluated by using direct substitution. So, in our case, we need that the
limit of π of π₯ as π₯ approaches negative two must be equal to π evaluated at
negative two. We can see from the question that
the function π evaluated at negative two is equal to five. So, to show that the function π in
our question is continuous, we must show that the limit of π of π₯ as π₯ approaches
negative two is equal to five.

So, we can now ask the question,
how do we evaluate the limit of π of π₯ as π₯ approaches negative two. We know from the definition of a
limit that the limit of a function as π₯ approaches π is equal to πΏ when both the
left-hand and right-hand limit of that function is equal to πΏ. So, since weβre told our function
π is continuous, and we know that the limit of π of π₯ as π₯ approaches negative
two is equal to five, we must also have that the limit of π of π₯ as π₯ approaches
negative two from the left is equal to five. And that the limit of π of π₯ as
π₯ approaches negative two from the right is equal to five.

So, letβs start with our limit of
π of π₯ as π₯ approaches negative two from the left, which we know is equal to
five. We have that π₯ is approaching
negative two from the left. So, we must have the π₯ is less
than negative two. From the question, we can see that
our function π of π₯ is equal to five π plus ππ₯ squared whenever π₯ is less than
negative two. So, since these functions are
equal, when π₯ is less than negative two, we can replace the π of π₯ in our left
limit with this polynomial.

Since weβre now trying to evaluate
the limit of a quadratic, we can use direct substitution. So, our limit is equal to five π
plus π multiplied by negative two squared, which we can then simplify it to see
that five is equal to five π plus four π. We can now apply a similar method
to evaluate the limit of π of π₯ as π₯ approaches negative two from the right. Which again we know is equal to
five because the question tells us that the function π is continuous at π₯ is equal
to negative two.

We see that since π₯ is approaching
negative two from the right, we must have that π₯ is greater than negative two. From our question, we can see that
our function π of π₯ is equal to ππ₯ minus three π whenever π₯ is greater than
negative two. Since our function π of π₯ is
equal to ππ₯ minus three π whenever π₯ is greater than negative two, we can just
replace our π of π₯ in our limit with ππ₯ minus three π.

Now, we can evaluate the limit of a
linear function by substitution. This gives us that our limit is
equal to π multiplied by negative two minus three π, which we can simplify to
negative two π minus three π. This gives us two equations for π
and π, which we know must be true. So, we can solve these as
simultaneous equations. We can eliminate the π term by
multiplying the first equation by two and the second equation by five.

Simplifying the first equation
gives us that 10 is equal to 10π plus eight π. And simplifying the second equation
will give us that 25 is equal to negative 10π minus 15π. We then add these two equations
together, giving us that 10 plus 25 is equal to 10π plus eight π added to negative
10π minus 15π. We can then simplify this to get
that 35 is equal to negative seven π. Which we can then solve to give us
that π is equal to negative five.

We can then substitute our value of
π equals negative five into the first equation. This would give us that five is
equal to five π plus four multiplied by negative five. Which we can solve to see that π
is equal to five. We can then substitute these values
of π equals five and π equals negative five into the definition of our function π
of π₯ to see that π of π₯ is equal to five multiplied by five plus negative five
multiplied by π₯ squared if π₯ is less than negative two. π of π₯ is equal to five if π₯ is
equal to negative two. And π of π₯ is equal to five π₯
minus three multiplied by negative five if π₯ is greater than negative two.

We can then simplify the piecewise
parts of our function π of π₯ to get a new definition of π of π₯. So, letβs clear some space and
discuss the differentiability of our function π at the point where π₯ is equal to
negative two. We can differentiate each part of
our piecewise function separately, where we ignore the part where π of π₯ is equal
to five, since this is only defined for a single point.

This gives us the following
piecewise definition for our derivative function, π prime of π₯. We can use the power rule to
evaluate each of these, giving us that our derivative function π prime of π₯ is
equal to negative 10π₯ if π₯ is less than negative two and five if π₯ is greater
than negative two. Finally, we need to check that the
left limit and the right limit as π₯ approaches negative two of π prime of π₯
agree. If the left and right limit agree,
we can say our function is differentiable at the point π₯ is equal to negative
two. If they do not agree, then we can
conclude that the function is not differentiable at the point π₯ is equal to
negative two.

For our right-hand limit, as π₯ is
approaching negative two from the right, we must have that π₯ is greater than
negative two. So, we must have that π prime of
π₯ is equal to negative 10π₯, giving us the limit of negative 10π₯ as π₯ approaches
negative two from the right. Since negative 10π₯ is a linear
function, we can evaluate this by direct substitution, giving us negative 10
multiplied by negative two. Which we can calculate to just be
20.

Similarly, for the limit as π₯
approaches negative two from the left, we must have that π₯ is less than negative
two. So, our function, π prime of π₯,
is equal to five, giving us the limit of five as π₯ approaches negative two from the
left. We know that for any constant π,
the limit of π as π₯ approaches π is equal to π. So, we can evaluate our limit of
five to just be equal to five. We can plot a graph at this point
to get a better idea of what we have shown.

We have the graphs of the piecewise
function π of π₯, which meet when π₯ is equal to negative two. The reason we get a corner here is
that our graph of π¦ equals five π₯ minus 15 has a constant gradient of five. However, the graph of π¦ equals 25
minus five π₯ squared has a gradient which approaches 20 at negative two. The difference between these two
values shows that our function π is not differentiable at the point π₯ is equal to
negative two.

So, to summarize what we have
found, we know that π is equal to five, π is equal to negative five, and the
function π of π₯ is not differentiable at the point π₯ is equal to negative
two.