Question Video: Finding the Unknown Coefficients in a Piecewise Function Given That the Function Is Differentiable at a Given Point Mathematics

Video Transcript

Find the values of 𝑎 and 𝑏 given the function 𝑓 is differentiable at 𝑥 is equal to negative one where 𝑓 of 𝑥 is equal to nine 𝑥 plus five if 𝑥 is less than negative one and 𝑓 of 𝑥 is equal to 𝑎𝑥 squared plus 𝑏𝑥 minus four if 𝑥 is greater than or equal to negative one.

We’re given a piecewise-defined function 𝑓 of 𝑥. We’re told that this piecewise-defined function is differentiable when 𝑥 is equal to negative one. We need to use this to determine the values of 𝑎 and 𝑏. And there’s a few different ways of doing this. We could do this directly from the definition of the derivative of a function when 𝑥 is equal to negative one. But there’s a simpler method by using our definition of the function 𝑓 of 𝑥. In this case, we can see our function 𝑓 of 𝑥 is defined piecewise. We can see the first piece is nine 𝑥 plus five and the second piece is 𝑎𝑥 cubed plus 𝑏𝑥 minus four.

In other words, in both cases, our function 𝑓 of 𝑥 is defined by polynomials which we know are continuous for all real values of 𝑥. When a piecewise-defined function is defined by continuous function, we call it piecewise continuous. And in fact, there’s something even more useful. We know how to differentiate polynomial functions by using the power rule for differentiation. We could use this to find an expression for 𝑓 prime of 𝑥, everywhere except where 𝑥 is equal to negative one. Then, we can use this for an easier method to check that our function 𝑓 of 𝑥 differentiable when 𝑥 is equal to negative one.

We just need to check that our function is continuous when 𝑥 is equal to negative one and the slope as 𝑥 approaches negative one from the left of 𝑓 of 𝑥 is equal to the slope as 𝑥 approaches negative one from the right of 𝑓 of 𝑥. But in this case, we already know our function 𝑓 is differentiable when 𝑥 is equal to negative one. This gives us two pieces of information about our function 𝑓. First, if 𝑓 is differentiable when 𝑥 is equal to negative one, then it also must be continuous when 𝑥 is equal to negative one. Second, we’ll also use the piece of information that 𝑓 is differentiable at 𝑥 is equal to negative one.

Let’s start with our first piece of information, 𝑓 is continuous when 𝑥 is equal to negative one. We know 𝑓 is a piecewise continuous function. And in fact, we can see 𝑥 is equal to negative one is the endpoint of our interval. And a piecewise continuous function can only be continuous at the endpoints of its interval if the endpoints match up. So, for our function 𝑓, the endpoints must match up. We could do this formally by using our left-hand and right-hand limit. In other words, we would check the limit as 𝑥 approaches negative one from the left of 𝑓 of 𝑥 is equal to the limit as 𝑥 approaches negative one from the right of 𝑓 of 𝑥.

But remember, we’ve already said that 𝑓 of 𝑥 is a piecewise continuous function. So, we could just evaluate these limits directly by substituting 𝑥 is equal to negative one into both of our functions. Because these are continuous, we can just evaluate by direct substitution. This gives us nine times negative one plus five is equal to 𝑎 times negative one all squared plus 𝑏 times negative one minus four. And we can simplify this expression.

Nine times negative one plus five simplifies to give us negative four and then eight times negative one squared plus 𝑏 times negative one minus four simplifies to give us 𝑎 minus 𝑏 minus four. And then, we can just add four to both sides of this equation, and this gives us zero is equal to 𝑎 minus 𝑏. If we add 𝑏 to both sides of this equation, we can see that 𝑎 must be equal to 𝑏. So because 𝑓 must be continuous at negative one, we’ve shown that 𝑎 must be equal to 𝑏.

We now want to use the fact that our function 𝑓 of 𝑥 must be differentiable when 𝑥 is equal to negative one. We could do this by directly using the definition of a derivative at 𝑥 is equal to negative one. However, 𝑓 of 𝑥 is defined in terms of polynomials, and we know how to differentiate polynomials by using the power rule for differentiation. So, instead of directly using the definition of a derivative when 𝑥 is equal to negative one, instead, we can use the fact that because 𝑓 must be differential when 𝑥 is equal to negative one, we must have the slope as 𝑥 approaches negative one from the left is equal to the slope as 𝑥 approaches negative one from the right.

And we can in fact evaluate both of these by just using the power rule for differentiation. So, we’ll use this to find an expression for 𝑓 prime of 𝑥. We get 𝑓 prime of 𝑥 is equal to the derivative of nine 𝑥 plus five with respect to 𝑥 if 𝑥 is less than negative one and 𝑓 prime of 𝑥 is equal to the derivative of 𝑎𝑥 squared plus 𝑎𝑥 minus four with respect to 𝑥 if 𝑥 is greater than negative one. And it’s important to know we don’t know what happens when 𝑥 is equal to negative one.

And it’s also important to point out since we know that 𝑎 is equal to 𝑏, we’ve just substituted this directly into our expression. Next, we need to evaluate both of these derivatives by using the power rule for differentiation. We need to multiply by the exponent of 𝑥 and then reduce this exponent by one. In fact, in our first function, we have a linear function. So, its slope will just be the coefficient of 𝑥, which is nine. Differentiating our second piece term by term by using the power rule for differentiation, we get two 𝑎𝑥 plus 𝑎 if 𝑥 is greater than negative one.

Now, we can use this to find the slope of 𝑓 of 𝑥 as 𝑥 approaches negative one from the left and the slope of 𝑓 of 𝑥 as 𝑥 approaches negative one from the right. Let’s start with the slope of 𝑓 of 𝑥 as 𝑥 approaches negative one from the left. Since our values of 𝑥 are approaching negative one from the left. We know our values of 𝑥 are less than negative one. So, our slope is just equal to the constant nine. So, our first expression simplifies to give us nine. We can do the same to find the slope as 𝑥 approaches negative one from the right.

Since 𝑥 is approaching negative one from the right, our values of 𝑥 will be greater than negative one. And we know the slope of our function when 𝑥 is greater than negative one is equal to two 𝑎𝑥 plus 𝑎. And, of course, two 𝑎𝑥 plus 𝑎 is a linear function. So, we can evaluate this limit by using direct substitution. We substitute 𝑥 is equal to negative one into this expression. This gives us two 𝑎 times negative one plus 𝑎, which of course we can simplify to give us negative two 𝑎 plus 𝑎 which is equal to negative 𝑎. But remember, we’re told our function 𝑓 is differentiable when 𝑥 is equal to negative one.

So, the slope as 𝑥 approaches negative one from the left of 𝑓 of 𝑥 must be equal to the slope as 𝑥 approaches negative one from the right of 𝑓 of 𝑥. So, we can just equate these values. We get negative 𝑎 must be equal to nine. And if we multiply through by negative one, we see that 𝑎 is equal to negative nine. And for completion, remember, we showed that 𝑎 must be equal to 𝑏. So, we can also say that 𝑏 must be equal to negative nine.

Therefore, if the function 𝑓 of 𝑥 is equal to nine 𝑥 plus five of 𝑥 is less than negative one and 𝑓 of 𝑥 is equal to 𝑎𝑥 squared plus 𝑏𝑥 minus four if 𝑥 is greater than or equal to negative one is differentiable when 𝑥 is equal to negative one, then we’ve shown that the value of 𝑎 must be equal to negative nine and the value of 𝑏 must also be equal to negative nine.