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.
