Video: Finding the Unknown Coefficients in a Quadratic Function given the Minimum Value of the Function

Given that the function 𝑓(π‘₯) = π‘₯Β² + 𝐿π‘₯ +𝑀 has a minimum value of 2 at π‘₯ = βˆ’1, determine the values of 𝐿 and 𝑀.

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Video Transcript

Given that the function 𝑓 of π‘₯ equals π‘₯ squared plus 𝐿 π‘₯ plus 𝑀 has a minimum value of two at π‘₯ 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 π‘₯ 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 π‘₯.

A minimum is a type of critical point. And we recall then that, at critical points, the gradient of the function 𝑓 prime of π‘₯ is equal to zero. We can use the power rule to differentiate 𝑓 of π‘₯. And we have that 𝑓 prime of π‘₯ is equal to two π‘₯ plus 𝐿. 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 𝑓 prime of π‘₯, 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 π‘₯ equals negative one. So when π‘₯ equals negative one, 𝑓 of π‘₯ is equal to two. So we can substitute negative one for π‘₯, two for 𝐿, and two for 𝑓 of π‘₯ 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 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 𝑓 prime of π‘₯ was two π‘₯ plus 𝐿. But we now know that 𝐿 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.

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.

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