Video: Finding the Critical Point of a Cubic Function in a Given Interval

Determine the critical points of the function 𝑦 = βˆ’8π‘₯Β³ in the interval [βˆ’2, 1].

03:11

Video Transcript

Determine the critical points of the function 𝑦 equals negative eight π‘₯ cubed in the interval negative two, one.

First, we recall that, at the critical points of a function, the gradient d𝑦 by dπ‘₯ will be equal to zero. So we need to find the gradient function for this curve. We can apply the power rule of differentiation. And it tells us that d𝑦 by dπ‘₯ is equal to negative eight multiplied by three multiplied by π‘₯ squared, which simplifies to negative 24π‘₯ squared.

Next, we set our expression for d𝑦 by dπ‘₯ equal to zero, giving the equation negative 24π‘₯ squared equals zero. We now need to solve for π‘₯. And we can see that as negative 24 isn’t equal to zero, it must be the case that π‘₯ squared is equal to zero. And if π‘₯ squared is equal to zero, then π‘₯ itself must be equal to zero. So we found the value of π‘₯ at our critical point.

Now in the question, we were asked to determine the critical points only in a particular interval, the interval negative two to one. And our value of π‘₯ does lie in this interval. In fact, it is the only critical point of this function.

Next, we need to find the value of 𝑦 at this critical point, which we can do by substituting our π‘₯-value back into the equation of the function. The function was 𝑦 equals negative eight π‘₯ cubed. So we have 𝑦 equals negative eight multiplied by zero cubed, which is just zero. So the only critical point in this interval and in fact the only critical point for the entire function has coordinates zero, zero. We can also see this if we sketch a graph of 𝑦 equals negative eight π‘₯ cubed.

Now this is the graph of 𝑦 equals π‘₯ cubed, which we should be familiar with. And it has a critical point. In fact, it’s a point of inflection at the origin. Multiplying by eight will cause a stretch with a scale factor of eight in the vertical direction. But this doesn’t affect the critical point. And then multiplying by negative one will cause a reflection in the π‘₯-axis. So in green, we have the graph of 𝑦 equals negative eight π‘₯ cubed. We can see that it does indeed have a critical point a point of inflection at the origin.

So by first finding the gradient function d𝑦 by dπ‘₯ and then setting it equal to zero and solving the resulting equation, we found the π‘₯-coordinate of our critical point. We then substituted this back into the equation of the original function in order to find the corresponding 𝑦-coordinate, giving a critical point of zero, zero. In our next example, we’ll consider how to determine the type of critical point without needing to sketch a graph.

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