### Video Transcript

Given that the function π of
π₯ equals π₯ squared plus πΏ π₯ plus π has a minimum value of two at π₯ equals
negative one, determine the values of πΏ and π.

In this question, weβve been
told the minimum value of the function. It is two. And weβve been told the value
of π₯ at which this occurs. Itβs negative one. We need to use this information
to calculate the missing coefficients πΏ and π in the definition of π of
π₯.

A minimum is a type of critical
point. And we recall then that, at
critical points, the gradient of the function π prime of π₯ is equal to
zero. We can use the power rule to
differentiate π of π₯. And we have that π prime of π₯
is equal to two π₯ plus πΏ. As negative one is the π₯-value
at a critical point, we know that if we substitute π₯ equals negative one into
our expression for π prime of π₯, we must get a result of zero. So we can form an equation. Two multiplied by negative one
plus πΏ equals zero. This gives the equation
negative two plus πΏ equals zero, which we can solve to give πΏ equals two.

So we found the value of
πΏ. But what about the value of
π? Well, we know that the function
has a minimum value of two when π₯ equals negative one. So when π₯ equals negative one,
π of π₯ is equal to two. So we can substitute negative
one for π₯, two for πΏ, and two for π of π₯ to give a second equation. Negative one squared plus two
multiplied by negative one plus π is equal to two. This simplifies to one minus
two plus π is equal to two, which again we can solve to give π is equal to
three.

We found the values of πΏ and
π. πΏ is equal to two, and π is
equal to three. But letβs just confirm that
this point is indeed a minimum point. We can do this using the
first-derivative test. We evaluate the first
derivative, π prime of π₯, either side of our critical value of π₯ so that we
can see how the gradient of this function is changing around the critical
point.

Remember, at the critical point
itself, the gradient is equal to zero. Our gradient function π prime
of π₯ was two π₯ plus πΏ. But we now know that πΏ is
equal to two. So our gradient function is two
π₯ plus two. When π₯ is equal to negative
two, this will give negative two. And when π₯ is equal to zero,
this will give positive two.

Itβs actually the sign of the
value rather than the value itself that weβre interested in. We see that the gradient is
negative to the left of negative one. Itβs then zero at negative one
itself and then positive to the right of negative one. So by sketching this pattern,
we see that the critical point at negative one is indeed a local minimum.

We finished the problem
then. πΏ is equal to two, and π is
equal to three.