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

Let π(π₯) = 5π +ππ₯Β², if π₯ < β2, and 5, if π₯ = β2, and ππ₯ β 3π, if π₯ > β2. Determine the values of π and π so that π is continuous at π₯ = β2. What can be said of the differentiability of π at this point?

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### Video Transcript

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

The question is asking us to first determine the values of π and π which make the function π continuous at the point π₯ is equal to negative two. The question then wants us to check if the function π 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 π is continuous at a point π if the limit of the function as π₯ approaches π can be evaluated by using direct substitution. So, in our case, we need that the limit of π of π₯ as π₯ approaches negative two must be equal to π evaluated at negative two. We can see from the question that the function π evaluated at negative two is equal to five. So, to show that the function π in our question is continuous, we must show that the limit of π 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 π 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 is equal to πΏ when both the left-hand and right-hand limit of that function is equal to πΏ. So, since weβre told our function π is continuous, and we know that the limit of π of π₯ as π₯ approaches negative two is equal to five, we must also have that the limit of π of π₯ as π₯ approaches negative two from the left is equal to five. And that the limit of π of π₯ as π₯ approaches negative two from the right is equal to five.

So, letβs start with our limit of π 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 π of π₯ is equal to five π plus ππ₯ 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 π of π₯ in our left limit with this polynomial.

Since weβre now trying to evaluate the limit of a quadratic, we can use direct substitution. So, our limit is equal to five π plus π multiplied by negative two squared, which we can then simplify it to see that five is equal to five π plus four π. We can now apply a similar method to evaluate the limit of π 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 π 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 π of π₯ is equal to ππ₯ minus three π whenever π₯ is greater than negative two. Since our function π of π₯ is equal to ππ₯ minus three π whenever π₯ is greater than negative two, we can just replace our π of π₯ in our limit with ππ₯ minus three π.

Now, we can evaluate the limit of a linear function by substitution. This gives us that our limit is equal to π multiplied by negative two minus three π, which we can simplify to negative two π minus three π. This gives us two equations for π and π, which we know must be true. So, we can solve these as simultaneous equations. We can eliminate the π 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 plus eight π. And simplifying the second equation will give us that 25 is equal to negative 10π minus 15π. We then add these two equations together, giving us that 10 plus 25 is equal to 10π plus eight π added to negative 10π minus 15π. We can then simplify this to get that 35 is equal to negative seven π. Which we can then solve to give us that π is equal to negative five.

We can then substitute our value of π equals negative five into the first equation. This would give us that five is equal to five π plus four multiplied by negative five. Which we can solve to see that π is equal to five. We can then substitute these values of π equals five and π equals negative five into the definition of our function π of π₯ to see that π of π₯ is equal to five multiplied by five plus negative five multiplied by π₯ squared if π₯ is less than negative two. π of π₯ is equal to five if π₯ is equal to negative two. And π 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 π of π₯ to get a new definition of π of π₯. So, letβs clear some space and discuss the differentiability of our function π 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 π 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, π prime of π₯. We can use the power rule to evaluate each of these, giving us that our derivative function π 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 π 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 π 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, π 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 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 π of π₯, which meet when π₯ is equal to negative two. The reason we get a corner here is that our graph of π¦ equals five π₯ minus 15 has a constant gradient of five. However, the graph of π¦ 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 π is not differentiable at the point π₯ is equal to negative two.

So, to summarize what we have found, we know that π is equal to five, π is equal to negative five, and the function π of π₯ is not differentiable at the point π₯ is equal to negative two.