Video Transcript
Given that the function 𝑓 of
𝑥 equals 𝑥 squared plus 𝐿 𝑥 plus 𝑀 has a minimum value of two at 𝑥 equals
negative one, determine the values of 𝐿 and 𝑀.
In this question, we’ve been
told the minimum value of the function. It is two. And we’ve been told the value
of 𝑥 at which this occurs. It’s negative one. We need to use this information
to calculate the missing coefficients 𝐿 and 𝑀 in the definition of 𝑓 of
𝑥.
A minimum is a type of critical
point. And we recall then that, at
critical points, the gradient of the function 𝑓 prime of 𝑥 is equal to
zero. We can use the power rule to
differentiate 𝑓 of 𝑥. And we have that 𝑓 prime of 𝑥
is equal to two 𝑥 plus 𝐿. As negative one is the 𝑥-value
at a critical point, we know that if we substitute 𝑥 equals negative one into
our expression for 𝑓 prime of 𝑥, we must get a result of zero. So we can form an equation. Two multiplied by negative one
plus 𝐿 equals zero. This gives the equation
negative two plus 𝐿 equals zero, which we can solve to give 𝐿 equals two.
So we found the value of
𝐿. But what about the value of
𝑀? Well, we know that the function
has a minimum value of two when 𝑥 equals negative one. So when 𝑥 equals negative one,
𝑓 of 𝑥 is equal to two. So we can substitute negative
one for 𝑥, two for 𝐿, and two for 𝑓 of 𝑥 to give a second equation. Negative one squared plus two
multiplied by negative one plus 𝑀 is equal to two. This simplifies to one minus
two plus 𝑀 is equal to two, which again we can solve to give 𝑀 is equal to
three.
We found the values of 𝐿 and
𝑀. 𝐿 is equal to two, and 𝑀 is
equal to three. But let’s just confirm that
this point is indeed a minimum point. We can do this using the
first-derivative test. We evaluate the first
derivative, 𝑓 prime of 𝑥, either side of our critical value of 𝑥 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 𝑓 prime
of 𝑥 was two 𝑥 plus 𝐿. But we now know that 𝐿 is
equal to two. So our gradient function is two
𝑥 plus two. When 𝑥 is equal to negative
two, this will give negative two. And when 𝑥 is equal to zero,
this will give positive two.
It’s actually the sign of the
value rather than the value itself that we’re interested in. We see that the gradient is
negative to the left of negative one. It’s 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. 𝐿 is equal to two, and 𝑀 is
equal to three.
