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

# Question Video: Finding the Local Maximum and Minimum Values of a Polynomial Function Mathematics • Third Year of Secondary School

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Determine the local maximum and minimum values of the function 𝑦 = 2𝑥³ + 6𝑥² − 11.

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### Video Transcript

Determine the local maximum and minimum values of the function 𝑦 equals two 𝑥 cubed plus six 𝑥 squared minus 11.

So, the first thing we need to consider is what are local maximum and local minimum points. Well, local maximum and local minimum points, as I’ve shown here on the sketch of any general function, are the points where our slope is equal to zero. We see that cause they’re the turning points on the function. So, if they’re where the slope is equal to zero, that means the first thing we need to do is find the slope function of our function and then set it equal to zero to help us find out what the 𝑥- and 𝑦-values are at these points.

So, if we differentiate two 𝑥 cubed plus six 𝑥 squared minus 11, we’re gonna get our slope function. So, we’re gonna get d𝑦 d𝑥. So, when we do that, we get six 𝑥 squared plus 12𝑥. And just to remind us how we differentiated, what we did is we multiplied the exponent by the coefficient. So, three multiplied by two gave us a six. And then, we reduce the exponent by one, so three minus one gives us a two. So, we get six 𝑥 squared. And then, we did the same for the second term. And then the final term, just cause we’ve got a constant, if we differentiate negative 11, we just get zero. So, great, now we’ve got our slope function.

So, now if we set it equal to zero, we can find our critical points, or zero values, where 𝑥 is going to be at these points. So, now to solve to find 𝑥, what we do is we factor the expression. So, we can take out six 𝑥 cause six 𝑥 is a factor of both six 𝑥 squared and 12𝑥. And inside the parentheses, we’re gonna have 𝑥 plus two.

And we got those two values because we need either six 𝑥 or 𝑥 plus two to equal zero to make the whole result zero. Well, to make six 𝑥 equal zero, if 𝑥 is zero then six 𝑥 is zero. And then, to make 𝑥 plus two equal to zero, it’s got to be negative two for our 𝑥-value because negative two plus two is zero. And then we’ve got six 𝑥 multiplied by zero, which would just be zero.

So, now to find our 𝑦-values, cause what we’re trying to do is actually find the local maximum and minimum values, so we want to know what the value of the function is at these points, we’re gonna substitute in 𝑥 equals zero and negative two into 𝑦 equals two 𝑥 cubed plus six 𝑥 squared minus 11. So, first of all, we’re gonna start with 𝑥 equals zero. So, we’re gonna get 𝑦 equals two multiplied by zero cubed plus six multiplied by zero squared minus 11, which is gonna give us a 𝑦-value, or a value of the function, of negative 11.

Then, next what we’re gonna do is substitute in 𝑥 is equals negative two. So, when we do that, we’re gonna get two multiplied by negative two cubed plus six multiplied by negative two squared minus 11, which means we’re gonna get negative 16 because we had two multiplied by negative two all cubed, where negative two cubed is negative eight. So, we get negative 16 plus 24 minus 11. So, this is gonna give us a value of the function as negative three.

So therefore, we found our local maximum and minimum values. But what we need to do is determine whether they are in fact maximum or minimum values. And to do that, we’re gonna use the second derivative.

Well, to find our second derivative, what we need to do is differentiate our slope function. So, that’s differentiate six 𝑥 squared plus 12𝑥. And when we do that, we get 12𝑥 plus 12. Okay, great, we’ve found that but why is it useful? Well, it’s useful because if we substitute in our values for 𝑥 that we found earlier, then we can determine whether our points are maximum or minimum values.

And that’s because if our second derivative is greater than zero, then it’s going to be a minimum value. If it’s less than zero, then it’s gonna be a maximum value. So, what we’re going to do now is substitute in 𝑥 equals zero and negative two into 12𝑥 plus 12. When we do that, we get 12 multiplied by zero plus 12, which is just going to give us a value of 12. So, we know that this is going to be a minimum value because this is greater than zero.

However, if we substitute in negative two, we’re gonna get 12 multiplied by negative two which is negative 24 plus 12 gives us negative 12. So therefore, this is going to be a local maximum because it is less than zero. So therefore, we can say that the function 𝑦 equals two 𝑥 cubed plus six 𝑥 squared minus 11 has a local minimum at negative 11 and a local maximum at negative three.

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