Determine the critical points of
the function 𝑦 equals negative eight 𝑥 cubed in the interval negative two,
First, we recall that, at the
critical points of a function, the gradient d𝑦 by d𝑥 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 d𝑥 is
equal to negative eight multiplied by three multiplied by 𝑥 squared, which
simplifies to negative 24𝑥 squared.
Next, we set our expression for d𝑦
by d𝑥 equal to zero, giving the equation negative 24𝑥 squared equals zero. We now need to solve for 𝑥. And we can see that as negative 24
isn’t equal to zero, it must be the case that 𝑥 squared is equal to zero. And if 𝑥 squared is equal to zero,
then 𝑥 itself must be equal to zero. So we found the value of 𝑥 at our
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 𝑥 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 𝑥-value back into
the equation of the function. The function was 𝑦 equals negative
eight 𝑥 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 𝑥 cubed.
Now this is the graph of 𝑦 equals
𝑥 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 𝑥-axis. So in green, we have the graph of
𝑦 equals negative eight 𝑥 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 d𝑥 and then setting it equal to zero and solving the resulting
equation, we found the 𝑥-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.