Video: Discussing the Continuity and Differentiability of a Piecewise Function at a Point

Discuss the continuity and differentiability of the function 𝑓 at π‘₯ = 0 given 𝑓(π‘₯) = βˆ’9π‘₯ βˆ’ 6, if π‘₯ < 0 and 𝑓(π‘₯) = π‘₯Β² βˆ’ 9π‘₯ βˆ’ 6, if π‘₯ β‰₯ 0.

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Video Transcript

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

We’re given a piecewise function defined as 𝑓 of π‘₯ is negative nine π‘₯ minus six if π‘₯ is less than zero and 𝑓 of π‘₯ is π‘₯ squared minus nine π‘₯ minus six if π‘₯ is greater than or equal to zero. Let’s first sketch the functions 𝑦 is negative nine π‘₯ minus six and 𝑦 is π‘₯ squared minus nine π‘₯ minus six on the same graph to get an idea of what we’re looking at.

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

We’re asked to discuss the continuity and differentiability of our function 𝑓 of π‘₯ at π‘₯ is equal to zero. So let’s define what we mean by continuity and differentiability and go from there.

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

We’re asked to discuss the continuity and differentiability of the function at π‘₯ is equal to zero. So let’s 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, 𝑓 of π‘₯ is 𝑓 of zero, which is zero squared minus nine times zero minus six. And that’s equal to negative six. So our first condition for continuity is satisfied. 𝑓 of zero exists, and it’s equal to negative six.

Now let’s check our second condition for continuity. Does the limit as π‘₯ tends to zero of 𝑓 of π‘₯ exist? We know that the definition of our function differs either side of π‘₯ is equal to zero. So we’re going to need to consider the behavior of 𝑓 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 that’s 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 that’s 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 𝑓 of π‘₯ is equal to 𝑓 of zero, which is negative six, our third condition for continuity is also satisfied. We can therefore say that 𝑓 of π‘₯ is continuous at π‘₯ is equal to zero.

So now let’s discuss the differentiability of our function at π‘₯ is equal to zero. Remember that a function 𝑓 of π‘₯ is differentiable at π‘₯ is equal to π‘Ž if 𝑓 prime of π‘Ž exists. Remember that 𝑓 prime of π‘₯ is d𝑓 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 π‘Žπ‘₯ to the power 𝑛 is equal to π‘Žπ‘›π‘₯ to the 𝑛 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.

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

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