Question Video: Finding the Local Maximum and Minimum Values of a Polynomial Function Mathematics

Video Transcript

Find the local maximum and minimum values of 𝑓𝑥 is equal to negative two 𝑥 cubed plus three 𝑥 squared plus 12𝑥.

Now what I’ve done is a little sketch just to kind of identify what we’re gonna be doing in this question. So we want to actually find the local maximum and minimums of our function. So I’ve drawn a sketch here of our function. And as you can see, at the maximum and minimum points, our slope or 𝑚 is gonna be equal to zero. So in order to actually find where these points are, we’re actually going to differentiate our function, negative two 𝑥 cubed plus three 𝑥 squared plus 12𝑥, in order to find the slope function.

So if we differentiate our function, our first term is going to be negative six 𝑥 squared. Just to remind us, we got that because we multiplied the exponent by the coefficient, so three multiplied by negative two, which gives negative six. And now we reduce the exponent by one. So we got 𝑥 squared.

Okay, great! Our next term will just be positive six 𝑥. And our final term will be positive 12, to leave us with the slope function, that is, negative six 𝑥 squared plus six 𝑥 plus 12. Okay, so if we look back at the sketch we did, we can see that, to find the maximum and minimum points, we need to find our points where our slope is equal to zero. So in order to do that, we’ve actually made our slope function zero. So we’ve now got zero is equal to negative six 𝑥 squared plus six 𝑥 plus 12.

So now we can start to solve this. So first of all, what we’re gonna do is actually multiply each side by negative one. That’s just to give us a positive 𝑥 squared term. So we get zero is equal to six 𝑥 squared minus six 𝑥 minus 12. I’ve just done that step just cause it makes it easier to solve at this point.

Now we’re actually gonna divide three by six. And we’re gonna do that again cause it simplifies, it makes our quadratic easier to solve. So we’re now left with zero is equal to 𝑥 squared minus 𝑥 minus two, which we can now solve by factoring, because if we factor our quadratic, what we get is 𝑥 minus two multiplied by 𝑥 plus one. And we achieve that because if we look at negative two and positive one, well the product needs to be negative two. So negative two multiplied by one gives us negative two. And the sum needs to be equal to negative one, because that’s the coefficient of our 𝑥 term. Well, negative two add one is negative one.

Okay, great! So now we have our factors. Let’s use them to find our values. So then we get 𝑥 is equal to two or negative one. And what these are, these are actually our 𝑥-coordinates of our critical values. So now what we want to do is we actually want to find out what the value of our function will be at these points. So I’m gonna substitute in 𝑥 is equal to two and 𝑥 is equal to negative one.

So if we substitute 𝑥 is equal to two into our function, it’s gonna be equal to negative two multiplied by two cubed plus three multiplied by two squared plus 12 multiplied by two, which gives us negative 16 plus 12 plus 24, which equals 20. So, therefore, one of our points is two, 20. So that’s gonna be one of our critical points. So it’s actually gonna be a maximum or minimum.

So now when we substitute in negative one for 𝑥, we get negative two multiplied by negative one cubed plus three multiplied by negative one squared plus 12 multiplied by negative one. So we’re gonna get two plus three minus 12, which gives us a value of negative seven. So, therefore, our next point is negative one, negative seven.

Okay, great! So we’ve got our two points. But now I need to find out which is the local maximum and which one is the local minimum. So in order to decide whether it’s in local maximum or minimum, we can actually use this relationship about the second derivative to actually help us, cause we know that if the second derivative is greater than zero, then it’s gonna be a local minimum. But if the second derivative is less than zero, then it’s gonna be a local maximum.

And we know that because actually if we look at the second derivative, if it’s gonna be positive or greater than zero, then that’s gonna be looking at the part where the graph is, in fact, concave up. So it’s gonna be our minimum point. If it’s less than zero or negative, then it’s actually gonna become concave down.

Okay, great! Now we know this. Let’s find the second derivative. So in order to find the second derivative, what we’re gonna do is we’re actually gonna differentiate our slope function, which was negative six 𝑥 squared plus six 𝑥 plus 12. And when we do that, we’re gonna be left with negative 12𝑥 plus six. So now what we’re gonna do is actually gonna substitute in our values of 𝑥 to determine whether it’s actually local maximum or local minimum points.

So first of all, if we substitute 𝑥 is equal to two, we’re gonna get negative 12 multiplied by two plus six, which is gonna be equal to negative 24 plus six, which is equal to negative 18. So, therefore, this is gonna be the local maximum point. And if we substitute in negative one for 𝑥, we’re gonna get negative 12 multiplied by negative one plus six, which is equal to 12 plus six, which is equal to 18. And as 18 is positive, therefore, this is gonna be our local minimum point.

Okay, great! We can therefore say that the local minimum value of 𝑓𝑥 is equal to negative two 𝑥 cubed plus three 𝑥 squared plus 12𝑥 is equal to negative seven at 𝑥 is equal to negative one. And our local maximum value is equal to 20 at 𝑥 is equal to two.