Video Transcript
Determine, if any, the local
maximum and minimum values of 𝑓 of 𝑥 equals negative two 𝑥 cubed minus nine
𝑥 squared minus 12𝑥 minus 15 together with where they occur.
In this question, we were asked
to determine the local maximum and minimum values. So that’s the values of the
function itself. And “with where they occur”
means we also need to find the corresponding 𝑥-values. First, we recall that, at the
critical points of a function, the gradient — so that’s 𝑓 prime of 𝑥 — is
equal to zero. So we begin by differentiating
𝑓 of 𝑥, which we can do using the power rule, to give negative six 𝑥 squared
minus 18𝑥 minus 12. At critical points, 𝑓 prime of
𝑥 is equal to zero. So we set our expression for 𝑓
prime of 𝑥 equal to zero.
And we’ll now solve the
resulting equation for 𝑥. We can factor negative six from
this equation, giving negative six multiplied by 𝑥 squared plus three 𝑥 plus
two is equal to zero. And we see that, inside the
parentheses, we have a quadratic in 𝑥 which does factorise. It’s equal to 𝑥 plus two
multiplied by 𝑥 plus one. We also note at this point that
negative six is not equal to zero. So we can eliminate it from our
equation at this point.
We now set each factor in turn
equal to zero, giving 𝑥 plus two equals zero or 𝑥 plus one equals zero. Both equations can be solved in
a relatively straightforward way to give the 𝑥-coordinates of the critical
points of this function. 𝑥 is equal to negative two or
𝑥 is equal to negative one.
We now know the 𝑥-values at
our critical points. But we also need to know the
values of the function itself. So we need to evaluate 𝑓 of 𝑥
at each critical value. When 𝑥 is equal to negative
two, 𝑓 of negative two is negative two multiplied by negative two cubed minus
nine multiplied by negative two squared minus 12 multiplied by negative two
minus 15, which is equal to negative 11. Evaluating 𝑓 of negative one
in the same way gives negative 10.
So we’ve now established that
this function has critical points at negative two, negative 11 and negative one,
negative 10. But how do we determine whether
they’re local maxima or minima or indeed points of inflection? Well, we’re going to use
something called the first-derivative test. We’ll consider the sign of the
derivative either side of our critical point, which will tell us the gradient of
the function either side of the critical point. By considering this, we’ll be
able to identify the shape of the function near to each critical point.
Here’s what we’re going to
do. We’re going to evaluate the
first derivative — that’s 𝑓 prime of 𝑥 — a little bit either side of our
critical 𝑥-values of negative two and negative one. Now, usually, we try to just
use the nearest integer values. But in this case, negative two
and negative one are consecutive integers. So instead, we’ve chosen a
value between them to be the upper value for negative two and the lower value
for negative one. We’ve chosen a value of
negative 1.5.
Remember that our gradient
function 𝑓 prime of 𝑥 is negative six 𝑥 squared minus 18𝑥 minus 12. So when we evaluate this at
negative three, we get a value of negative 12. Now we’re not particularly
interested in what the value is, but rather its sign. So negative 12 is a negative
value. We’ll also evaluate our
gradient function at negative 1.5. And it gives 1.5, which is a
positive value.
Finally, we need to evaluate
this gradient function at zero. And it gives negative 12, a
negative value. So how does this help us with
determining whether the critical points are maxima or minima? Well, we see that the gradient
of this curve is negative when 𝑥 is equal to negative three. It’s then equal to zero when 𝑥
equals negative two, and it’s positive when 𝑥 equals negative 1.5. And by sketching that shape, we
see that the critical point at negative two must be a local minimum.
In the same way, the gradient
of this function is positive when 𝑥 equals negative 1.5. It’s zero when 𝑥 equals
negative one. And it’s negative when 𝑥
equals zero. So we see that the critical
point at 𝑥 equals negative one must be a local maximum. So this method, the
first-derivative test, we consider the first derivative — that’s the gradient —
either side of the critical point. And by considering the sign of
this gradient, we can deduce the shape of the curve at this point. We found then that this
function, 𝑓 of 𝑥, has a local minimum at negative two, negative 11 and a local
maximum at negative one, negative 10.
