Question Video: Finding the Limit of a Function from Its Graph at a Point of Removable Discontinuity If the Limit Exists Mathematics

Video Transcript

Which of the following is the slope field of the differential equation 𝑦 prime is equal to two 𝑥 plus three 𝑦 minus five.

The way we’re going to approach this is by a process of elimination. First, we’ll find a point on each graph, where 𝑦 prime, the slope, is equal to zero. That’s a flat line segment. If our differential equation does not equal to zero at such a point, then we can eliminate the associated graph. If we start with graph A, d𝑦 by d𝑥 is equal to zero at the point one, negative one. So the slope in graph A is equal to zero at the point one, negative one. Now, if we substitute 𝑥 is equal to one and 𝑦 is negative one in our differential equation, we have 𝑦 prime is equal to two times one plus three times negative one minus five. That’s equal to two minus three minus five, which is negative six. This is nonzero. So straight away, we can eliminate graph A.

Now let’s look at graph B. Graph B has a slope of zero when 𝑥 is negative one and 𝑦 is negative one. If we try this in our equation, we have 𝑦 prime is equal to two times negative one plus three times negative one minus five. That’s equal to negative two minus three minus five, which is equal to negative 10. This is nonzero, so we can now eliminate graph B. Now let’s look at graph C. One of the points in graph C, where d𝑦 by d𝑥 is equal to zero, is the point three, two. With 𝑥 is three and 𝑦 is two in our equation, we have 𝑦 prime is two times three plus three times two minus five. That’s equal to seven, which is nonzero. And so we can eliminate equation C. Remember, we’re trying to find a graph with the point where the slope is equal to zero and where that point makes our differential equation also equal to zero.

Let’s look at graph D. The slope is equal to zero in graph D at the point one, one. Substituting 𝑥 equal to one and 𝑦 equal to one into our differential equation, we have 𝑦 prime is equal to two times one plus three times one minus five. That’s equal to two plus three minus five, which is equal to zero. This matches with the slope on the graph at the point one, one. So graph D is a contender. Now, let’s look at graph E. We can see that, in graph E, at the point one, one, we have a slope of zero. We know already from graph D that the point one, one satisfies our differential equation. So let’s try another point on graph E which has a slope of zero. The slope is equal to zero in graph E at the point four, minus one. So let’s try this in our differential equation. We have 𝑦 prime is two times four plus three times negative one minus five. That’s eight minus three minus five, which is equal to zero. So far then, graph E also matches our differential equation.

We now have two contenders, graph D and graph E. So let’s look again at graph D. In graph D, the slope is equal to zero at the point one, one. And this matches with our differential equation. The slope is also equal to zero at the point three, negative two. So let’s try this point in our differential equation. We have 𝑦 prime is two times three plus three times negative two minus five. That gives us six minus six minus five. And that’s equal to negative five, which is not zero. So we can now eliminate graph D. The only graph out of the five that could possibly match our differential equation now is graph E.

Let’s just try another couple of points in graph E with slope of zero and see if they match our differential equation. The slope on the graph of zero at the point four, negative one and in fact 𝑦 prime in the differential equation is also equal to zero. The slope on the graph is equal to zero at the point negative two, three. And 𝑦 prime is also equal to zero at the point negative two, three. Out of our five graphs then, we can say that graph E is the only graph that matches the differential equation 𝑦 prime is two 𝑥 plus three 𝑦 minus five.