Video: Finding Unknowns in a Piecewise-Defined Function That Make It Continuous at a Point Then Checking Its Differentiability at This Point

Let 𝑓(π‘₯) = 5π‘Ž +𝑏π‘₯Β², if π‘₯ < βˆ’2, and 5, if π‘₯ = βˆ’2, and π‘Žπ‘₯ βˆ’ 3𝑏, if π‘₯ > βˆ’2. Determine the values of π‘Ž and 𝑏 so that 𝑓 is continuous at π‘₯ = βˆ’2. What can be said of the differentiability of 𝑓 at this point?

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Video Transcript

Let 𝑓 of π‘₯ be the piecewise-defined function which is equal to five π‘Ž plus 𝑏π‘₯ squared if π‘₯ is less than negative two. 𝑓 of π‘₯ is equal to five when π‘₯ is equal to negative two. And 𝑓 of π‘₯ is equal to π‘Žπ‘₯ minus three 𝑏 when π‘₯ is greater than negative two. Determine the values of π‘Ž and 𝑏 so that 𝑓 is continuous at π‘₯ is equal to negative two. What can be said of the differentiability of 𝑓 at this point?

The question is asking us to first determine the values of π‘Ž and 𝑏 which make the function 𝑓 continuous at the point π‘₯ is equal to negative two. The question then wants us to check if the function 𝑓 will be differentiable at the point π‘₯ is equal the negative two. We can start by asking what makes a function continuous at a point.

We recall that a function 𝑓 is continuous at a point π‘Ž if the limit of the function as π‘₯ approaches π‘Ž can be evaluated by using direct substitution. So, in our case, we need that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two must be equal to 𝑓 evaluated at negative two. We can see from the question that the function 𝑓 evaluated at negative two is equal to five. So, to show that the function 𝑓 in our question is continuous, we must show that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is equal to five.

So, we can now ask the question, how do we evaluate the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two. We know from the definition of a limit that the limit of a function as π‘₯ approaches π‘Ž is equal to 𝐿 when both the left-hand and right-hand limit of that function is equal to 𝐿. So, since we’re told our function 𝑓 is continuous, and we know that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is equal to five, we must also have that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two from the left is equal to five. And that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two from the right is equal to five.

So, let’s start with our limit of 𝑓 of π‘₯ as π‘₯ approaches negative two from the left, which we know is equal to five. We have that π‘₯ is approaching negative two from the left. So, we must have the π‘₯ is less than negative two. From the question, we can see that our function 𝑓 of π‘₯ is equal to five π‘Ž plus 𝑏π‘₯ squared whenever π‘₯ is less than negative two. So, since these functions are equal, when π‘₯ is less than negative two, we can replace the 𝑓 of π‘₯ in our left limit with this polynomial.

Since we’re now trying to evaluate the limit of a quadratic, we can use direct substitution. So, our limit is equal to five π‘Ž plus 𝑏 multiplied by negative two squared, which we can then simplify it to see that five is equal to five π‘Ž plus four 𝑏. We can now apply a similar method to evaluate the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two from the right. Which again we know is equal to five because the question tells us that the function 𝑓 is continuous at π‘₯ is equal to negative two.

We see that since π‘₯ is approaching negative two from the right, we must have that π‘₯ is greater than negative two. From our question, we can see that our function 𝑓 of π‘₯ is equal to π‘Žπ‘₯ minus three 𝑏 whenever π‘₯ is greater than negative two. Since our function 𝑓 of π‘₯ is equal to π‘Žπ‘₯ minus three 𝑏 whenever π‘₯ is greater than negative two, we can just replace our 𝑓 of π‘₯ in our limit with π‘Žπ‘₯ minus three 𝑏.

Now, we can evaluate the limit of a linear function by substitution. This gives us that our limit is equal to π‘Ž multiplied by negative two minus three 𝑏, which we can simplify to negative two π‘Ž minus three 𝑏. This gives us two equations for π‘Ž and 𝑏, which we know must be true. So, we can solve these as simultaneous equations. We can eliminate the π‘Ž term by multiplying the first equation by two and the second equation by five.

Simplifying the first equation gives us that 10 is equal to 10π‘Ž plus eight 𝑏. And simplifying the second equation will give us that 25 is equal to negative 10π‘Ž minus 15𝑏. We then add these two equations together, giving us that 10 plus 25 is equal to 10π‘Ž plus eight 𝑏 added to negative 10π‘Ž minus 15𝑏. We can then simplify this to get that 35 is equal to negative seven 𝑏. Which we can then solve to give us that 𝑏 is equal to negative five.

We can then substitute our value of 𝑏 equals negative five into the first equation. This would give us that five is equal to five π‘Ž plus four multiplied by negative five. Which we can solve to see that π‘Ž is equal to five. We can then substitute these values of π‘Ž equals five and 𝑏 equals negative five into the definition of our function 𝑓 of π‘₯ to see that 𝑓 of π‘₯ is equal to five multiplied by five plus negative five multiplied by π‘₯ squared if π‘₯ is less than negative two. 𝑓 of π‘₯ is equal to five if π‘₯ is equal to negative two. And 𝑓 of π‘₯ is equal to five π‘₯ minus three multiplied by negative five if π‘₯ is greater than negative two.

We can then simplify the piecewise parts of our function 𝑓 of π‘₯ to get a new definition of 𝑓 of π‘₯. So, let’s clear some space and discuss the differentiability of our function 𝑓 at the point where π‘₯ is equal to negative two. We can differentiate each part of our piecewise function separately, where we ignore the part where 𝑓 of π‘₯ is equal to five, since this is only defined for a single point.

This gives us the following piecewise definition for our derivative function, 𝑓 prime of π‘₯. We can use the power rule to evaluate each of these, giving us that our derivative function 𝑓 prime of π‘₯ is equal to negative 10π‘₯ if π‘₯ is less than negative two and five if π‘₯ is greater than negative two. Finally, we need to check that the left limit and the right limit as π‘₯ approaches negative two of 𝑓 prime of π‘₯ agree. If the left and right limit agree, we can say our function is differentiable at the point π‘₯ is equal to negative two. If they do not agree, then we can conclude that the function is not differentiable at the point π‘₯ is equal to negative two.

For our right-hand limit, as π‘₯ is approaching negative two from the right, we must have that π‘₯ is greater than negative two. So, we must have that 𝑓 prime of π‘₯ is equal to negative 10π‘₯, giving us the limit of negative 10π‘₯ as π‘₯ approaches negative two from the right. Since negative 10π‘₯ is a linear function, we can evaluate this by direct substitution, giving us negative 10 multiplied by negative two. Which we can calculate to just be 20.

Similarly, for the limit as π‘₯ approaches negative two from the left, we must have that π‘₯ is less than negative two. So, our function, 𝑓 prime of π‘₯, is equal to five, giving us the limit of five as π‘₯ approaches negative two from the left. We know that for any constant π‘˜, the limit of π‘˜ as π‘₯ approaches π‘Ž is equal to π‘˜. So, we can evaluate our limit of five to just be equal to five. We can plot a graph at this point to get a better idea of what we have shown.

We have the graphs of the piecewise function 𝑓 of π‘₯, which meet when π‘₯ is equal to negative two. The reason we get a corner here is that our graph of 𝑦 equals five π‘₯ minus 15 has a constant gradient of five. However, the graph of 𝑦 equals 25 minus five π‘₯ squared has a gradient which approaches 20 at negative two. The difference between these two values shows that our function 𝑓 is not differentiable at the point π‘₯ is equal to negative two.

So, to summarize what we have found, we know that π‘Ž is equal to five, 𝑏 is equal to negative five, and the function 𝑓 of π‘₯ is not differentiable at the point π‘₯ is equal to negative two.

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