Video: ο»ΏDiscussing the Differentiability of a Piecewise-Defined Function at a Point

Suppose 𝑓(π‘₯) = βˆ’6π‘₯ βˆ’ 4, if π‘₯ ≀ βˆ’1 and 𝑓(π‘₯) = 3π‘₯Β², if π‘₯ > βˆ’1. What can be said about the differentiability of the function at π‘₯ = βˆ’1?

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Video Transcript

Suppose 𝑓 of π‘₯ is equal to negative six π‘₯ minus four, if π‘₯ is less than or equal to negative one, and three π‘₯ squared if π‘₯ is greater than negative one. What can be said about the differentiability of the function at π‘₯ equals negative one?

Here, we have been given a piecewise defined function composed of two subfunctions, a binomial and a monomial. Both of these subfunctions on their own or what’s known as smooth and will be defined over all the real numbers. In fact, on their own, we can say that all polynomials are smooth, which means that they’re differentiable on the real numbers. For a piecewise defined function, however, we must check the point at which are two subfunctions join. That is, in this case, π‘₯ equals negative one. To begin the process, let us differentiate both of our subfunctions using the power rule.

The derivative of negative six π‘₯ minus four is negative six times π‘₯ to the power of zero, which is, of course, just negative six. The derivative of three π‘₯ squared is two times three π‘₯, which is, of course, six π‘₯. We can represent this more succinctly by saying that 𝑓 dash of π‘₯ is equal to a negative six, if π‘₯ is less than negative one, and six π‘₯ if π‘₯ is greater than negative one. It should be noted that even though we see a less than or equal to inequality symbol here, we are not yet making any claims as to the value of our derivative when π‘₯ is equal to negative one, since this is what we’re trying to find at the moment.

Considering the derivative either side of the point where our two subfunctions join, we might try to move forward by substituting the value of π‘₯ equals negative one into the two subfunctions that we just found for 𝑓 dash of π‘₯. Doing so, we would find that both of our values are negative six. We may then conclude that since these two values are equal, that our function is differentiable at the point where π‘₯ is equal to negative one. Unfortunately, this is not the case. And to see why. Let us recall the definition of the derivative as a limit.

Here we’ve shown the limit which defines the derivative. To move forward, we recognize that since our function is defined differently on either side of the point where π‘₯ equals negative one, the left- and the right-sided limits will, therefore, have different expressions. Remembering that our π‘₯ nought is equal to negative one, we can begin by expressing the left limit as follows. Now, clearly, we cannot simply substitute β„Ž equals zero into this expression. Otherwise, we’ll be left with the indeterminate form of zero over zero. Instead, we know that when π‘₯ is less than negative one, 𝑓 of π‘₯ is equal to negative six π‘₯ minus four. Hence 𝑓 of negative one plus β„Ž will evaluate to two minus six β„Ž.

By similar reasoning, when π‘₯ is equal to negative one 𝑓 of π‘₯ takes the same subfunction. And therefore, 𝑓 of negative one evaluates to two substituting these in we find that the expression for our left-sided limit becomes two minus six β„Ž minus two all over β„Ž, which simplifies to negative six β„Ž over β„Ž. Now at this stage, since we know that β„Ž is approaching zero and not equal to zero, we’re allowed to cancel the common factor on the top and bottom half of our quotient. We are then left with the limit as β„Ž approaches zero from the left of negative six which, of course, is just equal to negative six.

Now for the right-sided limit, by similar reasoning as previously, we know that when π‘₯ is greater than negative one, 𝑓 of π‘₯ is defined by the subfunction three π‘₯ squared. Hence, in this case, 𝑓 of negative one plus β„Ž evaluates to three minus six β„Ž plus three β„Ž squared. For 𝑓 of negative one, we should be careful not to use our three π‘₯ squared again. Since when π‘₯ is equal to negative one, 𝑓 is defined by our other subfunction. We’ve already found the value of this to be two earlier. Again, we go through the same process of substituting these in and simplifying the expression for our limit. This time we reach a different result. Looking at the first term of our limit, we’ll have the limit as β„Ž approaches zero from the positive direction of one over β„Ž, which is equal to infinity. By extension, this means our entire right-sided limit is also equal to infinity.

We know that this is a particular way of expressing that the limit does not exist. If the right limit does not exist, then this also tells us that the normal limit does not exist. And hence, the derivative is not defind. The reason this is the case is that our graph actually has a discontinuity at π‘₯ equals negative one. And if we were to draw it, we would see this. Since we have concluded that the derivative is undefined at π‘₯ equals negative one, we’re also in a position to say that the function is not differentiable at π‘₯ equals negative one. And in fact, this is the answer to our question.

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