Video: Finding the Absolute Maximum and Minimum Values of a Polynomial Function in a Given Interval

Determine the absolute maximum and minimum values of the function 𝑦 = 2π‘₯Β³ + π‘₯Β² βˆ’ 3π‘₯ βˆ’ 2, in the interval [βˆ’1, 1], approximated to two decimal places.


Video Transcript

Determine the absolute maximum and minimum values of the function 𝑦 equals two π‘₯ cubed plus π‘₯ squared minus three π‘₯ minus two, in the interval negative one, one, approximated to two decimal places.

So we want to find the maximum and minimum values of our function. These values may be at the beginning of the interval, at the end, or somewhere in the middle. To do this problem, we’ll use a method called the closed interval method, which involves testing the function for a maximum and minimum at the endpoints and any turning points within the interval. Our curve is going to look something like this. And we can see there are potentially two turning points. Let’s go ahead with the closed interval method.

So this method will find the absolute maximum and minimum values of a continuous function on a closed interval. The first step is to find the critical numbers and evaluate 𝑓 at the critical numbers within the interval. The second step is to evaluate our function at the endpoints. Finally, the largest value obtained is the absolute maximum. And the smallest value obtained is the absolute minimum.

So let’s start with step one, finding the critical numbers and evaluating 𝑓 at these points. Let’s recall that critical numbers are the points where the slope is zero. And so, we find critical numbers by differentiating the function and setting it equal to zero, and then solving for π‘₯. We differentiate our function by firstly remembering that the derivative of π‘Žπ‘₯ to the power of 𝑛 is π‘›π‘Žπ‘₯ to the power of 𝑛 minus one. And so, d𝑦 by dπ‘₯ equals six π‘₯ squared plus two π‘₯ minus three. And we remember that constants differentiate to zero. So negative two just differentiates to zero. And as we said, we set this equal to zero and solve for π‘₯.

We can’t solve this by factoring. So let’s solve it by completing the square. We recall that for a quadratic in the form π‘₯ squared plus 𝑏π‘₯ plus 𝑐, the formula to complete the square is π‘₯ plus 𝑏 over two squared minus 𝑏 over two squared plus 𝑐. And it’s important to remember that when you complete the square, we want the π‘₯ squared coefficient to be one. So to do this, we’ll have to divide through by six. This would give us π‘₯ squared plus two over six π‘₯ minus three over six equals zero. But two over six is just a third, and three over six simplifies to a half. So we have that π‘₯ squared plus one-third π‘₯ minus one-half equals zero.

And now, we can apply the formula for completing the square. With 𝑏 being equal to one-third and 𝑐 being equal to negative one-half, we apply the formula for completing the square. And remember that when you divide a fraction by a number, you can just multiply the denominator of the fraction by the number. So one-third over two is just one-sixth. Now, remember that we’re aiming to eventually get π‘₯ on its own. So let’s add a half and a sixth squared to both sides. This gives us that π‘₯ plus one-sixth squared equals one-sixth squared plus a half. And one-sixth squared is just one-sixth multiplied by one-sixth. And fraction-multiplication rules tells us that this is going to be one over 36.

Let’s now tidy up the right-hand side by adding these two fractions together. We’ll do this by using the fact that one over two is equivalent to 18 over 36. And then one over 36 plus 18 over 36 is 19 over 36. Remember that we’re still solving for π‘₯. So let’s square root both sides. And we remember that our solution could be positive or negative. To simplify the right-hand side, we know from surd rules that the square root of 19 over 36 is the same as the square root of 19 over the square root of 36. And the square root of 36 is six. And finally, to isolate π‘₯, we can subtract one-sixth from both sides. So π‘₯ equals negative one-sixth plus or minus root 19 over six.

So here are our two solutions and we can write each of these as a single fraction. Putting it into a calculator, this gives us the values of π‘₯, 0.56 and negative 0.89, rounded to two decimal places. Going back to step one of the closed interval method, we can see that we need to substitute these values of π‘₯ back into our original function. We must be careful here to use the exact values rather than the rounded values, just to avoid any rounding errors. And we’re going ahead and evaluating our function at these points. Because both the values we got for π‘₯ are in our interval between negative one and one. So I’ll clear some space so that we can do this.

So if we substitute our first value of π‘₯ here into our function using a calculator, we get that 𝑦 equals negative 3.02, to two decimal places. And then substituting the other value we found for π‘₯, again using the exact value, we find that 𝑦 equals 0.05, to two decimal places. So step one is now complete.

Now step two tells us to evaluate 𝑓 at the interval endpoints. Our interval endpoints are negative one and one. When we substitute π‘₯ equals negative one into our function, we find that 𝑦 equals zero. And when we substitute π‘₯ equals one into our function, we find that 𝑦 equals negative two. So that’s step two complete. We’ve evaluated our function at the interval endpoints.

Finally, the last step says that the largest value obtained is the maximum and the smallest value is the minimum. So let’s have a look at the values that we obtained. 0.05 is the largest of these values. And negative 3.02 is the smallest value we obtained. Therefore, by the closed interval method, we have found that the absolute maximum of our function is 0.05. And the absolute minimum is negative 3.02.

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