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) = x²/₃(4 -x²) on -2,2

Accepted Answer

Hello. In this video, we are going to be identifying the absolute maximum and minimum values of the given function. The problem gives us the function F of X equal to 3 X raised to the power of 2/3, multiplied by the quantity 16 minus X2, and we are told that this function is defined on the interval from -4 to 4. In order to obtain the absolute maximum and minimum values, there are two pieces of information we need. First, we need to identify the values of the boundary points or the endpoints of the interval of the given function. That is given to us as -4 and 4, so we need to identify the value of F at these boundary points. We also need to identify the values of the function located at the critical points. Since we need to find the critical points, we will start this problem by finding the derivative of F of X. In order to take the derivative of the given function, we will go ahead and use the product rule. That will be F X equal to 16 minus X2 multiplied by the derivative of 3 X to the 2/3 power. That is going to be 2 multiplied by X raised to the negative 1/3. Then we will add 3 X to the 2/3 power multiplied by the derivative of 16 minus X2. That derivative is going to be negative to X. Next, we will go ahead and multiply 3 X to the 2/3 with -2 X. That will leave us with 16 minus X squared multiplied by 2 X to the negative 1/3. Minus 6 X raised to the power of 5/3, and this is going to be the derivative of the given function. Now that we have the derivative, what we want to do is find the critical points. Now, recall that the critical points exist when the derivative is equal to 0. So what we want to do with this derivative is set it equal to 0 and solve for X. Now, the first thing we want to do is you want to eliminate this negative exponent of X. We can eliminate this negative exponent by multiplying this equation through by X raise to the power of 1/3. That is going to leave us with 2 multiplied by the quantity 16 minus X squared, minus 6 X raise the power of 6/3, which will simplify to be 6 X squared, and this will equal to 0. Now what we will go ahead and do is simplify. We will distribute 2 into 16 minus X2. That will give us 32 minus 2 X2 minus 6 X2 equal to 0. -2 X 2 and -6X2 will combine to give us -8X2d. So we have 32 minus 8 X2 equal to 0. We will subtract 32 to both sides of the equation, leaving us with -8 X2 equal to -32. Dividing through by -8, we'll simplify this further to be X2 equal to 4. And finally, if we take the square root of both sides of the equation, we end up with X's equal to positive and negative 2. What this means is that there are two values for the critical points. There is a value at X, there is a critical point at 2, and a critical point at -2. So, we have two critical points and we have our boundaries of the interval -4 and 4. In order to identify the absolute extrema, we will take these values, plug them into the original function, and compare the values to each other. So, let's go ahead and start by plugging in -4 into the original function. F of -4 will equal to 3 multiplied by -4, raise the power of 2/3, multiplied by 16. Minus -4 2, and this is going to simplify to give us the value of 0. We will now plug in -2. F of -2 will equal to 3, multiplied by -2 rates of the 2/3, multiplied by the quantity 16 minus -2 squared. This will simplify to give us 36 multiplied by the cube root of 4, and this will approximate to be 57.15. We will plug in 2 to the original function, giving us 3 multiplied by 2, raise the power of 2/3, multiplied by 16 minus 2 squared. Which will simplify to be 36 multiplied by the cube root of 4, which again approximates to 57.15. And finally, when we plug in the value of 4, we get F of 4 equal to 3, multiplied by 4, raised to the power of 2/3, multiplied by 16 minus 4 squad, which will simplify to be 0. So, now that we have the values of the critical points, and we have the values of the endpoints of the original function, we can now compare the values to identify the absolute maximums and absolute minimums. Now notice that the critical points have the same value 57.15. And this value is greater than the values located at the endpoints of the interval. What this means is that there are 2 locations for absolute minimums. We have an absolute minimum or absolute maximum, I'm sorry. An absolute maximum value at F of 2, which has the same value of F of -2, which is equal. To 36 multiplied by the cube root of 4. And for the absolute minimum values, that is the least value amongst the four, which is both zero, and these values of the absolute minimums. are located at F of -4, which is equal to F4, which is equal to 0. And this is going to be the solution to this problem. So I hope this video helps you in helping you understand of how to find the absolute maximum minimum values, and I will go ahead and see you all in the next video.

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) = x²/₃(4 -x²) on -2,2

Key Concepts

Absolute Extrema

Critical Points

Endpoints of the Interval

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