Absolute maxima and minima Determine the location and value of...

Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = 2x³ - 15x² + 24x on [0,5]

Accepted Answer

Welcome back, everyone. On the interval open bracket negative to to closed bracket, determine the absolute maximum and minimum values of the function P X equals 4 X cubed minus 12 X2 + 8X. A says the absolute minimum is 0 and the absolute maximum is 96. B says the minimum is -1 and the maximum is 96. C says they are -1 and 193 respectively, and the D says they are negative 96 and 8 root 3 divided by 9 respectively. Now if we are going to determine the absolute maximum and minimum values of the function, we'll need the critical points of PX. And to find PX's critical points, we can set the first derivative of PX to 0 and the sulfur X. So at our critical points. Hm. Then we know the first derivative of PF X is going to be equal to zero. So this implies then that the first derivative of 4 X cubed minus 12 X2 + 8X, which when we differentiate would be 12 x quad minus 24 X plus 8, is going to be equal to 0. So no, we all we all we need to do now is to solve for X. Now, here we can solve this. Uh first, we can divide through by 4, sorry, to get our equation to be 3 X 2 minus 6 X plus 2. No, we can apply the quadratic formula because the coefficient of our squared term is 3, coefficient of our linear term is -6, and our constant is 2. So by the quadratic formula. OK. Then we can say that X is going to be equal to negative B, so that's 6 plus R minus the square root of B2-62. Minus 4 multiplied by A, multiplied by C. And now all of this is being divided by 2 multiplied by A. So that's 2 multiplied by 3. Now, if we do some work here, let me come over to the right. Then So this is gonna become 6, OK, plus R minus the square root of 12 divided by 6. Therefore, X is going to be equal to start that here. X is going to be equal to 6 + 2 root 3 divided by 6 or 6 minus 2 root 3 divided by 6. Now here 6 + 2 route 3 divided by 6 can be simplified to 1 + 3 of route 3, and on the other hand, 6 minus 2 route 3 divided by 6 will be 1 minus 3 of route 3. But no, the critical point X equals 1 plus route 3 divided by 3 does not lie in the interval between -2 to 2 inclusive. Therefore, OK. X equals 1 minus root 3, divided by 3, is the critical point. It's the critical point of our function. Now that we have the critical point, we can evaluate it along with the end points to figure out which one is the absolute maximum. So now to find the absolute maximum. And minimum So to find the absolute maximum and minimum. OK. We can substitute all our end points and critical points into P. and then we should be able to find it. So, for example, that means we're going to find P of -2, OK? P of 2. A P 1 minus root 3 divided by 3. Now, for P of -2, when X equals 2, then this is gonna be 4 multiplied by -2 cubed. OK. Minus 12 multiplied by -2 square plus 8 multiplied by -2, which equals -96. For p of 2, it's going to be 4 multiplied by 2 cubed minus 12 multiplied by 2 squared plus 8 multiplied by 2, which is going to be equal to 0. Finally, p 1 minus 3 of root 3 equals 4 multiplied by 1 minus 1/3 of root 3 cubed. Minus 12 multiplied by 1 minus 3 of root 3 squared plus 8 multiplied by 1 minus 3 of root 3. And now again, when we evaluate this, then we should get P to be equal to a root 3 divided by 9. So what does this tell us? Well, now, we know that the absolute minimum is -96, OK? And the absolute max sorry, not maximum, minimum. The absolute minimum is -96. My apologies, sorry, I think that. And the absolute maximum, OK, is 8 root 3 divided by 9. If we look back at our answer choices, that means D is the correct choice. Thanks for watching everyone. I hope this video helped.

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.ƒ(x) = 2x³ - 15x² + 24x on [0,5]

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Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = 2x³ - 15x² + 24x on [0,5]