# Question Video: Finding the Critical Point of a Cubic Function in a Given Interval Mathematics • 12th Grade

Determine the critical points of the function 𝑦 = −8𝑥³ in the interval [−2, 1].

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### Video Transcript

Determine the critical points of the function 𝑦 equals negative eight 𝑥 cubed in the interval negative two to one.

First, let’s recall what is meant by the term critical points. The critical points, also called stationary or turning points of a function, are points where the gradient of that function is equal to zero. The gradient may change sign from negative to positive, for example, as in our first diagram. Or it may have the same sign on either side of the critical point but be zero at the critical point itself.

We can find an expression for the gradient of a function at any point by differentiating that function. If the function is given as 𝑦 in terms of 𝑥, then we can find the derivative of 𝑦 with respect to 𝑥, d𝑦 by d𝑥. Our function is 𝑦 equals negative eight 𝑥 cubed. Let’s find its gradient function or derivative.

To differentiate a polynomial function of 𝑥, we bring the power down. So we have negative eight multiplied by three. We then decrease the power by one. So subtracting one from three gives two. Negative eight multiplied by three is negative 24. So the gradient function of this curve is d𝑦 by d𝑥 equals negative 24𝑥 squared.

Now as we’ve already said, critical points occur when the gradient at that point is equal to zero. So we’re going to set our gradient function d𝑦 by d𝑥 equal to zero and then solve the resulting equation, which will give us the 𝑥-coordinates of any critical points. Our gradient function is negative 24𝑥 squared. So setting this equal to zero gives the equation negative 24𝑥 squared is equal to zero.

We can see that as negative 24 isn’t equal to zero, 𝑥 squared must be equal to zero, so that when we multiply these two values together, we get zero. We could also see this by dividing both sides of the equation by negative 24. And as zero divided by negative 24 gives zero, we’ll have the equation 𝑥 squared is equal to zero.

To solve for 𝑥, we take the square root of each side of this equation. And as the square root of zero is just zero, we find that 𝑥 is equal to zero. So this is the 𝑥-coordinate at our critical point. And we see that it is in the interval negative two, one.

We now need to find the 𝑦-coordinate at our critical point, which we do by substituting this 𝑥-value back into the original function. 𝑦, remember, is equal to negative eight 𝑥 cubed. So substituting zero for 𝑥, we get that 𝑦 is equal to negative eight multiplied by zero cubed. Zero cubed is just zero. That’s zero times zero times zero. And multiplying by negative eight will still give zero. So we find that the 𝑦-coordinate of our critical point is also zero.

Our function only has one critical point, the point zero, zero. It is in the interval negative two, one. But it’s also the only critical point for this function.