Question Video: Finding Unknowns in a Piecewise-Defined Function That Make It Continuous at a Point Then Checking Its Differentiability at This Point Mathematics • Second Year of Secondary School

Video Transcript

Let 𝑓 of 𝑥 be the piecewise-defined function which is equal to five 𝑎 plus 𝑏𝑥 squared if 𝑥 is less than negative two. 𝑓 of 𝑥 is equal to five when 𝑥 is equal to negative two. And 𝑓 of 𝑥 is equal to 𝑎𝑥 minus three 𝑏 when 𝑥 is greater than negative two. Determine the values of 𝑎 and 𝑏 so that 𝑓 is continuous at 𝑥 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 𝑎 and 𝑏 which make the function 𝑓 continuous at the point 𝑥 is equal to negative two. The question then wants us to check if the function 𝑓 will be differentiable at the point 𝑥 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 𝑎 if the limit of the function as 𝑥 approaches 𝑎 can be evaluated by using direct substitution. So, in our case, we need that the limit of 𝑓 of 𝑥 as 𝑥 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 𝑥 as 𝑥 approaches negative two is equal to five.

So, we can now ask the question, how do we evaluate the limit of 𝑓 of 𝑥 as 𝑥 approaches negative two. We know from the definition of a limit that the limit of a function as 𝑥 approaches 𝑎 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 𝑥 as 𝑥 approaches negative two is equal to five, we must also have that the limit of 𝑓 of 𝑥 as 𝑥 approaches negative two from the left is equal to five. And that the limit of 𝑓 of 𝑥 as 𝑥 approaches negative two from the right is equal to five.

So, let’s start with our limit of 𝑓 of 𝑥 as 𝑥 approaches negative two from the left, which we know is equal to five. We have that 𝑥 is approaching negative two from the left. So, we must have the 𝑥 is less than negative two. From the question, we can see that our function 𝑓 of 𝑥 is equal to five 𝑎 plus 𝑏𝑥 squared whenever 𝑥 is less than negative two. So, since these functions are equal, when 𝑥 is less than negative two, we can replace the 𝑓 of 𝑥 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 𝑎 plus 𝑏 multiplied by negative two squared, which we can then simplify it to see that five is equal to five 𝑎 plus four 𝑏. We can now apply a similar method to evaluate the limit of 𝑓 of 𝑥 as 𝑥 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 𝑥 is equal to negative two.

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

Now, we can evaluate the limit of a linear function by substitution. This gives us that our limit is equal to 𝑎 multiplied by negative two minus three 𝑏, which we can simplify to negative two 𝑎 minus three 𝑏. This gives us two equations for 𝑎 and 𝑏, which we know must be true. So, we can solve these as simultaneous equations. We can eliminate the 𝑎 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 10𝑎 plus eight 𝑏. And simplifying the second equation will give us that 25 is equal to negative 10𝑎 minus 15𝑏. We then add these two equations together, giving us that 10 plus 25 is equal to 10𝑎 plus eight 𝑏 added to negative 10𝑎 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 𝑎 plus four multiplied by negative five. Which we can solve to see that 𝑎 is equal to five. We can then substitute these values of 𝑎 equals five and 𝑏 equals negative five into the definition of our function 𝑓 of 𝑥 to see that 𝑓 of 𝑥 is equal to five multiplied by five plus negative five multiplied by 𝑥 squared if 𝑥 is less than negative two. 𝑓 of 𝑥 is equal to five if 𝑥 is equal to negative two. And 𝑓 of 𝑥 is equal to five 𝑥 minus three multiplied by negative five if 𝑥 is greater than negative two.

We can then simplify the piecewise parts of our function 𝑓 of 𝑥 to get a new definition of 𝑓 of 𝑥. So, let’s clear some space and discuss the differentiability of our function 𝑓 at the point where 𝑥 is equal to negative two. We can differentiate each part of our piecewise function separately, where we ignore the part where 𝑓 of 𝑥 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 𝑥. We can use the power rule to evaluate each of these, giving us that our derivative function 𝑓 prime of 𝑥 is equal to negative 10𝑥 if 𝑥 is less than negative two and five if 𝑥 is greater than negative two. Finally, we need to check that the left limit and the right limit as 𝑥 approaches negative two of 𝑓 prime of 𝑥 agree. If the left and right limit agree, we can say our function is differentiable at the point 𝑥 is equal to negative two. If they do not agree, then we can conclude that the function is not differentiable at the point 𝑥 is equal to negative two.

For our right-hand limit, as 𝑥 is approaching negative two from the right, we must have that 𝑥 is greater than negative two. So, we must have that 𝑓 prime of 𝑥 is equal to negative 10𝑥, giving us the limit of negative 10𝑥 as 𝑥 approaches negative two from the right. Since negative 10𝑥 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 𝑥 approaches negative two from the left, we must have that 𝑥 is less than negative two. So, our function, 𝑓 prime of 𝑥, is equal to five, giving us the limit of five as 𝑥 approaches negative two from the left. We know that for any constant 𝑘, the limit of 𝑘 as 𝑥 approaches 𝑎 is equal to 𝑘. 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 𝑥, which meet when 𝑥 is equal to negative two. The reason we get a corner here is that our graph of 𝑦 equals five 𝑥 minus 15 has a constant gradient of five. However, the graph of 𝑦 equals 25 minus five 𝑥 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 𝑥 is equal to negative two.

So, to summarize what we have found, we know that 𝑎 is equal to five, 𝑏 is equal to negative five, and the function 𝑓 of 𝑥 is not differentiable at the point 𝑥 is equal to negative two.