### Video Transcript

Discuss the differentiability of
the function π of π₯ is equal to negative four π₯ plus one over π₯ at π₯ is equal
to negative seven.

Weβre given the function π of π₯
is negative four π₯ plus one over π₯. And weβre asked to discuss the
differentiability of this function at π₯ is negative seven.

The first thing to check when
considering the differentiability of a function at a point is whether or not the
function is continuous at that point. Thatβs because if a function π of
π₯ is differentiable at π₯ is equal to π, then π is continuous at π₯ is equal to
π. This means that for a function to
be differentiable at a point, it must be continuous at that point. And if it is not continuous at that
point, then it cannot be differentiable there. The converse of this statement is
not necessarily true though. A function can be continuous at a
point without being differentiable there.

So, we need our function to be
continuous at π₯ is equal to negative seven. Once weβve established this, weβll
determine whether the derivative exists at that point. There are three conditions that
must be satisfied for a function to be continuous at a point. These are: π at π₯ is equal to π
exists, the limit as π₯ tends to π of π of π₯ exists, and the limit as π₯ tends to
π of π of π₯ is equal to π at π₯ is equal to π.

Our function π of π₯ is negative
four π₯ plus one over π₯. Looking at the first condition, we
want to know whether π of negative seven exists or not. Substituting negative seven into
our function, we have negative four times negative seven plus one over negative
seven. Thatβs equal to 28 minus one over
seven, which evaluates to 24 over seven. So, π of negative seven does exist
and equals 24 over seven. Our first condition for continuity
is satisfied.

Now, we need to check the second
condition that the limit as π₯ tends to π of π of π₯ exists. In our case, that means checking
that the limit as π₯ tends to negative seven of π of π₯ exists. As π₯ approaches negative seven,
our function approaches negative four times negative seven plus one over negative
seven. We already know that thatβs 24 over
seven so that the limit as π₯ tends to negative seven of π of π₯ exists. And our second condition for
continuity at the point is satisfied.

For the third condition, we want to
know that the limit as π₯ tends to π of π of π₯ is π of π. We know that this is the case from
our first and second conditions. The limit as π₯ tends to negative
seven of π of π₯ is equal to 24 over seven, which is π of π. So, our third condition for
continuity is satisfied.

Since we know now that our function
is continuous at π₯ is negative seven, we could check for differentiability at that
point. Our condition for differentiability
at a point is π of π₯ is differentiable at π₯ is equal to π if π prime of π
exists. So, if we can show that the
derivative of π at π₯ is negative seven exists, then our function is differentiable
at that point.

We can write our derivative as π
prime of π₯ or dπ by dπ₯, which is Leibnizβs notation. And thatβs d by dπ₯ of negative
four π₯ plus one over π₯. Since one over π₯ is π₯ to the
negative one, thatβs d by dπ₯ of negative four π₯ plus π₯ to the negative one.

Differentiating our first term,
negative four π₯, we know that the derivative of π₯ multiplied by a constant is just
a constant. So, d by dπ₯ of negative four π₯ is
negative four. For our second term, π₯ to the
negative one, we can use the power rule. This says that d by dπ₯ of ππ₯ to
the π is equal to π times π times π₯ to the π minus one. In other words, we multiply by the
exponent and take one off the exponent. So, d by dπ₯ of π₯ to the negative
one is negative π₯ to the negative two, which is negative four minus one over π₯
squared.

Our question is: Does π prime of
negative seven exist? Substituting negative seven into π
prime, our dπ by dπ₯, we have negative four minus one over negative seven
squared. This evaluates to negative 4.02, to
two decimal places, so that our condition for differentiability that π prime of π
exists is satisfied. We can, therefore, say that π of
π₯ is equal to negative four π₯ plus one over π₯ is differentiable at π₯ is negative
seven because π prime at π₯ is negative seven exists.

Itβs worth noting that had we been
looking at π of π₯ at π₯ is equal to zero, the result wouldβve been different. Because we have the term one over
π₯ in our function at π₯ is equal to zero, this term is undefined. So, π of zero does not exist, and
the function is not continuous at that point and, therefore, not differentiable at
π₯ is equal to zero.