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) = x/(x²+9)⁵ on [-2,2]

Accepted Answer

Welcome back, everyone. In this problem, we want to find the absolute maximum and minimum values of the function H on the interval open bracket 04 close bracket where H of X equals 3x divided by X2 + 1 raises to the 4th. A says the absolute maximum is 1,029 Route 7 divided by 4,096 at X equals 0, and the absolute minimum is 0 at X equals 1 divided by Route 7. B says the absolute maximum is 0 at X equals 1 divided by Route 7, and the minimum is 1,029 route 7 divided by 4,0096 at X equals 0. C says the maximum is 0 at X equals 0 and 1,029 roots 7 divided by 4,0096 at X equals 1 divided by route 7, and D says the absolute maximum is 1,029, Route 7 divided by 4,0096, at X equals 1 divided by Route 7, and the minimum is 0 at X equals 0. Now, if we're going to find absolute maximum and minimum values, first we need to find a derivative of H of X. And we'll get to why we're finding it in a second, but for now, let's just focus on differentiating H of X. Notice here that H of X is a quotient so we can apply the quotient rule of differentiation. Recall, the quotient rule basically tells us that if we have a function Y that's equal to U divided by V where U and V are functions of X, then the derivative of Y will be equal to Vu minus UV divided by Vquared. Where U and V are derivatives of U and V respectively. So in this case, if we let our numerator be equal to U 3X and our denominator V be equal to X2 + 1 raise to the 4th, OK. Then notice right that the derivative of u will be equal to 3, while the derivative of V based on the chain rule will be equal to 8 X multiplied by X2 + 1 to the third. With that in mind Then differentiating H of X by the chain rule or by the quotient rule, the derivative is going to be equal to U multiplied by V. so that's 3 multiplied by X2 + 1 to the fourth minus U multiplied by V. So that's 3 X multiplied by 8 X multiplied by X2 + 1 to the 3, OK. I know we're gonna divide all of this by X2 + 1 to the 4th, to the part of 2, which is gonna be X2 + 1 to the 8th. I know if we simplify this expression, OK. Then notice here we can factor out X2 + 1 cubed from both terms, multiply 3 x by 8 X. We're going to end up with 3 multiplied by X2 + 1 minus 24 X 2 divided by X2 + 1 raised to the core of 1/5. Now, we can simplify that even further if we expand our bracket here, and we're gonna get 3 minus 21 X 2 divided by X2 + 1 to the 5th. This is the derivative of H of X. Now why do we need the derivative? Let's get back to that question. Well, at the critical points, OK, let me separate this with a line here just to show that we're working over here. At the critical points. critical point. OK. Our first derivative will be equal to 0. So what does that mean? We can set our derivative to 0 and solve for X to get the X values of our critical points. OK. In that case, we can say that 3 minus 21 X2 divided by X2 + 1 race to the 5th equals 0. Multiply through by our denominator, we're going to be left to 3 minus 21 X2 equals 0, OK. Thus, if we divide through by 3 and add 1 to both sides, then we're gonna get. And then divide through by 7, sorry, we're gonna get X squad to be equal to 1/7, which means that X will be equal to plus or minus the square root of 1/7. Remember that our interval is 0 to 4 inclusive, so that means. That we can take X to be positive route 1/7, OK. Uh, let me just put that in bracket here since our interval is 04. Or 0 to 4. Now, with that in mind, OK, here we have the X value for a critical point. So now, we can evaluate H OK, H of X at the critical points and endpoints to get our absolute maximum and minimum values. So that means essentially. We want to find the value of age. When X equals 0. When X equals root 1/7. And when X equals 4. No, when X equals 0 H of 0 is 0. When X equals route 1/7, H is going to be equal to 3 multiplied by route 1/7 divided by 17. That would be route 17 squared, which leaves us with just 1/7. Plus one. Raised to the power of 1/4 which would be equal to 1,029 Route 7 divided by 4,0096. Now finally, for our age of 4 is going to be equal to 12, 3 multiplied by 4 divided by 17 rates to the 4th, which will be equal to 12 divided by 83,521. No from our values of age here, which are going to be the maximum and the minimum. Well, notice here. That the maximum value. Is going to be 1,029 route 7 divided by 4,0096 at X equals route 17, OK? And the minimum value is going to be zero. That's our smallest value here at X equals 0. Thus the absolute maximum value is 1,029 Route 7 divided by 4,096 at X equals route 17, and the absolute minimum value is zero at X equals 0. If we look back at our answer choices, that tells us that D is the correct answer. Thanks a lot 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) = x/(x²+9)⁵ on [-2,2]

Key Concepts

Absolute Extrema

Critical Points

Closed Interval Method

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