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Determine the absolute maximum and minimum values of the function π¦ equals two π₯ cubed plus π₯ squared minus three π₯ minus two, in the interval negative one, one, approximated to two decimal places.
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So we want to find the maximum and minimum values of our function.
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These values may be at the beginning of the interval, at the end, or somewhere in the middle.
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To do this problem, weβll 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.
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Our curve is going to look something like this.
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And we can see there are potentially two turning points.
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Letβs go ahead with the closed interval method.
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So this method will find the absolute maximum and minimum values of a continuous function on a closed interval.
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The first step is to find the critical numbers and evaluate π at the critical numbers within the interval.
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The second step is to evaluate our function at the endpoints.
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Finally, the largest value obtained is the absolute maximum.
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And the smallest value obtained is the absolute minimum.
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So letβs start with step one, finding the critical numbers and evaluating π at these points.
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Letβs recall that critical numbers are the points where the slope is zero.
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And so, we find critical numbers by differentiating the function and setting it equal to zero, and then solving for π₯.
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We differentiate our function by firstly remembering that the derivative of ππ₯ to the power of π is πππ₯ to the power of π minus one.
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And so, dπ¦ by dπ₯ equals six π₯ squared plus two π₯ minus three.
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And we remember that constants differentiate to zero.
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So negative two just differentiates to zero.
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And as we said, we set this equal to zero and solve for π₯.
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We canβt solve this by factoring.
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So letβs solve it by completing the square.
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We recall that for a quadratic in the form π₯ squared plus ππ₯ plus π, the formula to complete the square is π₯ plus π over two squared minus π over two squared plus π.
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And itβs important to remember that when you complete the square, we want the π₯ squared coefficient to be one.
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So to do this, weβll have to divide through by six.
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This would give us π₯ squared plus two over six π₯ minus three over six equals zero.
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But two over six is just a third, and three over six simplifies to a half.
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So we have that π₯ squared plus one-third π₯ minus one-half equals zero.
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And now, we can apply the formula for completing the square.
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With π being equal to one-third and π being equal to negative one-half, we apply the formula for completing the square.
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And remember that when you divide a fraction by a number, you can just multiply the denominator of the fraction by the number.
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So one-third over two is just one-sixth.
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Now, remember that weβre aiming to eventually get π₯ on its own.
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So letβs add a half and a sixth squared to both sides.
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This gives us that π₯ plus one-sixth squared equals one-sixth squared plus a half.
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And one-sixth squared is just one-sixth multiplied by one-sixth.
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And fraction-multiplication rules tells us that this is going to be one over 36.
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Letβs now tidy up the right-hand side by adding these two fractions together.
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Weβll do this by using the fact that one over two is equivalent to 18 over 36.
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And then one over 36 plus 18 over 36 is 19 over 36.
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Remember that weβre still solving for π₯.
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So letβs square root both sides.
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And we remember that our solution could be positive or negative.
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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.
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And the square root of 36 is six.
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And finally, to isolate π₯, we can subtract one-sixth from both sides.
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So π₯ equals negative one-sixth plus or minus root 19 over six.
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So here are our two solutions and we can write each of these as a single fraction.
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Putting it into a calculator, this gives us the values of π₯, 0.56 and negative 0.89, rounded to two decimal places.
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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.
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We must be careful here to use the exact values rather than the rounded values, just to avoid any rounding errors.
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And weβre going ahead and evaluating our function at these points.
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Because both the values we got for π₯ are in our interval between negative one and one.
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So Iβll clear some space so that we can do this.
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So if we substitute our first value of π₯ here into our function using a calculator, we get that π¦ equals negative 3.02, to two decimal places.
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And then substituting the other value we found for π₯, again using the exact value, we find that π¦ equals 0.05, to two decimal places.
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So step one is now complete.
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Now step two tells us to evaluate π at the interval endpoints.
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Our interval endpoints are negative one and one.
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When we substitute π₯ equals negative one into our function, we find that π¦ equals zero.
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And when we substitute π₯ equals one into our function, we find that π¦ equals negative two.
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So thatβs step two complete.
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Weβve evaluated our function at the interval endpoints.
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Finally, the last step says that the largest value obtained is the maximum and the smallest value is the minimum.
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So letβs have a look at the values that we obtained.
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0.05 is the largest of these values.
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And negative 3.02 is the smallest value we obtained.
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Therefore, by the closed interval method, we have found that the absolute maximum of our function is 0.05.
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And the absolute minimum is negative 3.02.