Video: Discussing the Differentiability of a Function at a Point

Discuss the differentiability of the function 𝑓(π‘₯) = βˆ’4π‘₯ + (1/π‘₯) at π‘₯ = βˆ’7.

04:42

Video Transcript

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

We’re given the function 𝑓 of π‘₯ is negative four π‘₯ plus one over π‘₯. And we’re 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. That’s because if a function 𝑓 of π‘₯ is differentiable at π‘₯ is equal to π‘Ž, then 𝑓 is continuous at π‘₯ is equal to π‘Ž. 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 we’ve established this, we’ll 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: 𝑓 at π‘₯ is equal to π‘Ž exists, the limit as π‘₯ tends to π‘Ž of 𝑓 of π‘₯ exists, and the limit as π‘₯ tends to π‘Ž of 𝑓 of π‘₯ is equal to 𝑓 at π‘₯ is equal to π‘Ž.

Our function 𝑓 of π‘₯ is negative four π‘₯ plus one over π‘₯. Looking at the first condition, we want to know whether 𝑓 of negative seven exists or not. Substituting negative seven into our function, we have negative four times negative seven plus one over negative seven. That’s equal to 28 minus one over seven, which evaluates to 24 over seven. So, 𝑓 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‘Ž of 𝑓 of π‘₯ exists. In our case, that means checking that the limit as π‘₯ tends to negative seven of 𝑓 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 that’s 24 over seven so that the limit as π‘₯ tends to negative seven of 𝑓 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‘Ž of 𝑓 of π‘₯ is 𝑓 of π‘Ž. We know that this is the case from our first and second conditions. The limit as π‘₯ tends to negative seven of 𝑓 of π‘₯ is equal to 24 over seven, which is 𝑓 of π‘Ž. 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 𝑓 of π‘₯ is differentiable at π‘₯ is equal to π‘Ž if 𝑓 prime of π‘Ž exists. So, if we can show that the derivative of 𝑓 at π‘₯ is negative seven exists, then our function is differentiable at that point.

We can write our derivative as 𝑓 prime of π‘₯ or d𝑓 by dπ‘₯, which is Leibniz’s notation. And that’s d by dπ‘₯ of negative four π‘₯ plus one over π‘₯. Since one over π‘₯ is π‘₯ to the negative one, that’s 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 π‘Žπ‘₯ to the 𝑛 is equal to 𝑛 times π‘Ž times π‘₯ to the 𝑛 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 𝑓 prime of negative seven exist? Substituting negative seven into 𝑓 prime, our d𝑓 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 𝑓 prime of π‘Ž exists is satisfied. We can, therefore, say that 𝑓 of π‘₯ is equal to negative four π‘₯ plus one over π‘₯ is differentiable at π‘₯ is negative seven because 𝑓 prime at π‘₯ is negative seven exists.

It’s worth noting that had we been looking at 𝑓 of π‘₯ at π‘₯ is equal to zero, the result would’ve been different. Because we have the term one over π‘₯ in our function at π‘₯ is equal to zero, this term is undefined. So, 𝑓 of zero does not exist, and the function is not continuous at that point and, therefore, not differentiable at π‘₯ is equal to zero.

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