{Use of Tech} Absolute maxima and minima

a. F...

Question

{Use of Tech} Absolute maxima and minima

a. Find the critical points of f on the given interval.

b. Determine the absolute extreme values of f on the given interval.

c. Use a graphing utility to confirm your conclusions.

f(x) = 2ᶻ sin x on [-2,6]

Accepted Answer

Hello, in this video, we are going to be finding the critical points, the absolute maximum value, and the absolute minimum value of the function H of X is equal to 3 of X, multiplied by sin of X on the interval from -1 to 4. We want to round to 3 decimal places, and we also want to plot the function using a graphing utility. So Let's go ahead and just write down the function that is given to us. We are given the function H of X is equal to 3 of X, multiplied by sin of X, and we know that this function exists on the interval from -1 to 4. So, let's go ahead and approach the first part of this problem, which is finding the critical points of this function. In order to find the critical points, we want to find the values of X that cause its derivative to be zero. So, in order to find the critical points, the first thing we need to do is take the derivative of the function. We will go ahead and take the derivative by using the product rule. That will give us H X is equal to 3 of X, multiplied by the derivative of sine, which is cosine of X. Plus, sin of X, multiplied by the derivative of 3 power of X, which is 3 power of X, multiplied by the natural logarithm of 3. By multiplying terms together, we will end up with 3 to the power of X, cosine of X. Plus 3 to the power of X. Allen of 3 multiplied by sine of X. We can further simplify this function by factoring out a 3 to the power of X, and the derivative will be 3 power of X, multiplied by the quantity, cosine of X, plus the natural logarithm of sin of X. So now that we have the derivative, we're going to set the derivative equal to 0. And what we want to do now is we want to solve for the values of X. Now, we have two factors of X. If we set the first factor 3 the power of X equal to 0, notice that there is no value of X that we can plug into the exponential that will cause this exponential value to be zero. So, in this case here, there is no solution to this equation, therefore, we will disregard this factor of X. But now what we want to do is we want to find the values of X that cause cosine of X plus the natural logarithm of 3 sin of X to be zero. We will go ahead and solve this by using trigonometric identities. The first thing that we will do is we will go ahead and subtract sin of X, natural logarithm of 3, to the right hand side of the equation. That will leave us with cosine of X equal to the negative natural logarithm of 3, sine of X. What we are going to do next is we are going to multiply both sides of this equation, or divide both sides of this equation by cosine of X. If we divide both sides of this equation by cosine of X, we are left with 1 is equal to the negative natural logarithm of 3, multiplied by sin of x, divided by cosine of X. Then if we divide the negative natural logarithm of 3 to both sides of this equation, we are left with sin of X. Divided by cosine of X equal to -1 divided by the natural logarithm of 3. Notice that sin of x divided by cosine of X is equal to tangent of X. So we can rewrite the left hand side to be tangent of X, leaving us with tangent of X equal to the negative 1 divided by the natural logarithm of 3. Finally, we will go ahead and take the inverse tangent of both sides of this equation, leaving us with X's equal to tangent inverse of -1 divided by the natural logarithm of 3. And since this is a defined defined angle in the tangent inverse, we will also add the generic value k pi to represent the period of tangent. Furthermore, tangent of a negative angle can be rewritten with the negative sign moved in front of the tangent, since it is odd function. So we can rewrite this function one more time to be X is equal to the negative tangent inverse of 1 divided by the natural logarithm of 3, plus K pi. Now what we want to do is we want to find the values of X that are within the interval from -1 to 4. And in order to find these values, we are going to plug in different values of K. Let's go ahead and try out when K is equal to 0. When K is equal to 0, we get X is equal to the negative tangent inverse of 1 divided by the natural logarithm of 3 plus 0, and this is going to approximate to be negative 0.738. Since this is within our interval from -1 to 4, this is going to be the first critical point in the interval. Now we are going to repeat this process, but for K is equal to 2. When we have Ks equal to 2, we get X's equal to the negative tangent inverse of 1 divided by the natural logarithm of 3 plus 2 pi. And this is going to approximate to B. 5.5447, but this lies outside of the interval, so we will not be picking this value of K. But what about the value of K's equal to one? When K is equal to 1, we get X is equal to negative tangent inverse of 1 divided by the natural logarithm of 3 plus pi, and this is going to approximate to be 2.403, which is inside the interval from -1 to 4, and this is going to be the other critical point. So we have two critical points. We have -0.738. And we have 2.403. Now that we've identified the critical points, let's go ahead and move on to the second part of this equation, or this problem where it asks us to find the absolute maximum and absolute minimum values. So what we want to do is we want to go ahead and take our boundary values and the two critical points that we just calculated, and plug them into the original function. When we do this, we are going to compare the outputs to each other and determine which of these are going to be the absolute maximum or minimum values. So Let's go ahead and create some room here real quick. OK. So, what we are going to do is we are going to plug in the values of our boundaries, which is -1 to 4, into the original function. When we plug in -1 into the original function, we get the output of approximately 0.280. When we plug in our first critical point, we have the function with negative 0.738 plugged in, which will approximate to be negative 0.299. When we plug in our second critical point, the function with 2.403 plugged in, is going to approximate to give us 9.434. And finally, when we plug in our boundary.4 into the function, we get an approximate value of -61.301. So Which of these values is the largest value amongst the four? Well, that value is going to be at 2.403, with an output of 9.434. Since this is the largest output amongst the boundary points and the critical points, this is going to be the absolute maximum. Now which of these values is the smallest, has the smallest output amongst all four values? Well, that is going to be when we plug in 4 into the original function with an output of negative 61.301. This is going to be our absolute minimum. Now that we have the locations of the critical points and the boundary points, let's go ahead and plot these points on the on the graph. So, we have -1 at 0.280. So that is going to be Right below the X axis at -1. Then, we will go ahead and plot the first critical point, which is located at -0.7384. Or 0.738, which has an output of negative 0.299. So that's just going to be just above. Just to the left of the origin. Now, for the next critical point, we have an X value of 2.403 with an output of 9.434. That is going to be roughly. Here Halfway between 2 and 3 and just below 10. And for the boundary point at 4, we have an output of -61.301. So that's going to be located at 4 and just below -60. So when we plot the points together, we get this rough curve, which is going to be the approximate graph of the given function. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

{Use of Tech} Absolute maxima and minima

a. Find the critical points of f on the given interval.
b. Determine the absolute extreme values of f on the given interval.
c. Use a graphing utility to confirm your conclusions.

f(x) = 2ᶻ sin x on [-2,6]

Key Concepts

Critical Points

Absolute Extrema

Graphing Utility

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