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

Find the values of π‘Ž and 𝑏 given the function 𝑓 is differentiable at π‘₯ = 1 where 𝑓(π‘₯) = βˆ’π‘₯Β² + 4, if π‘₯ ≀ 1 and 𝑓(π‘₯) = βˆ’2π‘Žπ‘₯ βˆ’ 𝑏, if π‘₯ > 1.

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Video Transcript

Find the values of π‘Ž and 𝑏 given the function 𝑓 is differentiable at π‘₯ is equal to one where 𝑓 of π‘₯ is equal to negative π‘₯ squared plus four if π‘₯ is less than or equal to one and 𝑓 of π‘₯ is equal to negative two π‘Žπ‘₯ minus 𝑏 if π‘₯ is greater than one.

We’re given a piecewise-defined function 𝑓 of π‘₯. And we’re told that this function 𝑓 is differentiable when π‘₯ is equal to one. We need to use this to determine the values of π‘Ž and 𝑏. The first thing we notice about this is when π‘₯ is equal to one, we can see we’re at the endpoints of our interval of the piecewise-defined function. In other words, when π‘₯ is equal to one, our function 𝑓 of π‘₯ changes from being equal to negative π‘₯ squared plus four to being equal to negative two π‘Žπ‘₯ minus 𝑏.

And at this point, there’s a few different methods we could use to try and answer this question. For example, we might be tempted to directly use the definition of 𝑓 being differentiable at π‘₯ is equal to one. And this would work. However, because our function 𝑓 of π‘₯ is defined piecewise and π‘₯ is equal to one is one of the endpoints of this interval, we can actually do this in a simpler way.

First, we recall if a function is differentiable at some point, then it must also be continuous at this point. In other words, since we know 𝑓 is differentiable at π‘₯ is equal to one, we know that 𝑓 must also be continuous when π‘₯ is equal to one. And we can see something interesting about our function 𝑓 of π‘₯. We can see both pieces of this function are polynomials. And we know polynomials are continuous for all real values of π‘₯. So our function 𝑓 of π‘₯ is piecewise continuous.

And for a piecewise continuous function to be continuous at its endpoints, its endpoints must match. In other words, we know the limit as approaches one from the left of 𝑓 of π‘₯ must be equal to the limit as π‘₯ approaches one from the right of 𝑓 of π‘₯.

Now, we could evaluate this limit directly. However, we need to remember 𝑓 of π‘₯ is a piecewise continuous function. And because each piece is continuous, we can evaluate each of these limits by using direct substitution. So we just substitute π‘₯ is equal to one into negative π‘₯ squared plus four to evaluate the limit as π‘₯ approaches one from the left of 𝑓 of π‘₯. We get negative one squared plus four.

And we can do the same to evaluate the limit as π‘₯ approaches one from the right. We have negative two π‘Žπ‘₯ minus 𝑏 is continuous. So we can evaluate this limit by using direct substitution. We just substitute π‘₯ is equal to one into negative two π‘Žπ‘₯ minus 𝑏. This gives us negative two π‘Ž times one minus 𝑏. And because we know 𝑓 is continuous, we know these two limits have to be equal.

Let’s now simplify both sides of this equation. First, negative one squared plus four is equal to three. And we can simplify the right-hand side of this equation to give us negative two π‘Ž minus 𝑏. But this is only one equation with two variables. So we need more information to find the values of π‘Ž and 𝑏. To do this, we’ll want to specifically use the fact that 𝑓 is differentiable at π‘₯ is equal to one. And we know several different ways of explaining that 𝑓 is differentiable at π‘₯ is equal to one. But one of these is a lot easier to work with for our function 𝑓 of π‘₯.

We can see that both of the parts of 𝑓 of π‘₯ are defined as polynomials. And we already know how to differentiate polynomials term by term by using the power rule for differentiation. So instead of directly applying the definition of a derivative to our function 𝑓 of π‘₯, we can instead look at the slope as π‘₯ approaches one from the left of 𝑓 of π‘₯ and look at the slope as π‘₯ approaches one from the right of 𝑓 of π‘₯. In other words, we know if 𝑓 is differentiable at π‘₯ is equal to one, then the slope as π‘₯ approaches one from the left of 𝑓 of π‘₯ must be equal to the slope as 𝑓 approaches one from the right of 𝑓 of π‘₯. And we use this because we can easily find an expression for 𝑓 prime of π‘₯ when π‘₯ is less than one and when π‘₯ is greater than one.

We just need to differentiate each piece of 𝑓 of π‘₯ separately. We get 𝑓 prime of π‘₯ is equal to the derivative of negative π‘₯ squared plus four with respect to π‘₯ if π‘₯ is less than one and 𝑓 prime of π‘₯ is equal to the derivative of negative two π‘Žπ‘₯ minus 𝑏 with respect to π‘₯ if π‘₯ is greater than one. And it’s worth reiterating at this point we’re not stating what 𝑓 prime of π‘₯ is equal to when π‘₯ is equal to one. We’re just finding an expression for the slope for all of the other values of π‘₯.

Now, we can evaluate both of these derivatives by using the power rule for differentiation. We want to multiply by our exponents of π‘₯ and reduce this exponent by one. First, the derivative of negative π‘₯ squared plus four with respect to π‘₯ is equal to negative two π‘₯. Next, to differentiate our second function, we could again use the power rule for differentiation term by term. However, this is also a linear function. So we could just differentiate this by taking the coefficient of π‘₯, which is negative two π‘Ž. This gives us 𝑓 prime of π‘₯ is equal to negative two π‘₯ if π‘₯ is less than one and 𝑓 prime of π‘₯ is equal to negative two π‘Ž if π‘₯ is greater than one.

Now, we can evaluate the slope as π‘₯ approaches one from the left of 𝑓 of π‘₯ and the slope as π‘₯ approaches one from the right of 𝑓 of π‘₯. First, when π‘₯ approaches one from the left, we can see 𝑓 prime of π‘₯ is equal to negative two π‘₯. And of course, negative two π‘₯ is a continuous function. So we can evaluate this by using direct substitution. We just substitute π‘₯ is equal to one. We get negative two times one, which is equal to negative two.

And we can do exactly the same as π‘₯ approaches one from the right. This time, 𝑓 prime of π‘₯ will be equal to negative two π‘Ž. But this time, negative two π‘Ž is a constant. So this is just equal to negative two π‘Ž. And remember, we’re told that 𝑓 of π‘₯ is differentiable when π‘₯ is equal to one. So the slope as π‘₯ approaches one from the left of 𝑓 of π‘₯ must be equal to the slope as π‘₯ approaches one from the right of 𝑓 of π‘₯. In other words, we can equate these two values. We get negative two must be equal to negative two π‘Ž.

And if we divide both sides of this equation through by negative two, we see that π‘Ž must be equal to one. Now, to find our value 𝑏, we substitute π‘Ž is equal to one into our equation three is equal to negative two π‘Ž minus 𝑏. Substituting π‘Ž is equal to one gives us that three is equal to negative two times one minus 𝑏. And by simplifying and rearranging this equation, we can see that 𝑏 must be equal to negative five.

Therefore, if the function 𝑓 of π‘₯ is equal to negative π‘₯ squared plus four if π‘₯ is less than or equal to one and 𝑓 of π‘₯ is equal to negative two π‘Žπ‘₯ minus 𝑏 if π‘₯ is greater than one is differentiable when π‘₯ is equal to one, then we’ve shown that π‘Ž must be equal to one and 𝑏 must be equal to negative five.

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