Local max/min of x¹⸍ˣ Use analytical methods to find all local...

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Local max/min of x¹⸍ˣ Use analytical methods to find all local extrema of the function ƒ(x) = x¹⸍ˣ , for x > 0 . Verify your work using a graphing utility.

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Hello. In this video, we are going to be determining all the local extrema of F of X. The equation F of X is given to us as F of X equal to X rates the power of 3 divided by X. In order to find the local extrema, we want to find the values of X that cause the derivative to be zero, also known as the critical numbers. So, let's go ahead and start by taking the derivative of the given F of X function. Now, before we take the derivative, we can go ahead and simplify the function by using the natural logarithm properties. First, we will replace F of X with Y. We will have Y equal to X raise the power of 3 divided by X. In order to simplify using natural logarithms, we will first take the natural logarithm of both sides of the equation. That will leave us with natural logarithm of Y equal to the natural logarithm of X raise to the power of 3 divided by X. By using the power rule of logarithms, we can rewrite the right-hand side of the equation as 3 divided by X, multiplied by the natural logarithm of X. Now, what we can go ahead and do is we can go ahead and take the derivative with respect to X, but keep in mind that this is going to involve the derivative of the natural logarithm function. The derivative of a natural logarithm of X is defined as 1 divided by X. So using this property, let's go ahead and take the derivative of the equation with respect to X. On the left hand side, the derivative of the natural logarithm of Y will be 1 divided by Y. But since we are taking the derivative of Y with respect to X, we will generate a DYDX term through implicit differentiation. On the right-hand side, we are taking the derivative of 3 divided by X, multiplied by the natural logarithm of X. We will need to use the power rule with respect to X. That is going to give us 3 divided by X, multiplied by the derivative of the natural logarithm of X, which is 1 divided by X. Then we will add the natural logarithm of X, multiplied by the derivative of 3 divided by X, and that will be -3 divided by X2d through the quotient rule. Now we will go ahead and simplify the derivative. On the right hand side, multiplying 3 divided by X with 1 divided by X will give us 3 divided by X squared. And multiplying the natural logarithm of X with -3 divided by X2 will give us negative 3 natural logarithm of X divided by X2. We can further simplify the right-hand side of the equation by combining both of the fractions into a single fraction. That will give us 3 minus 3 natural logarithm of X divided by X2, and that is equal to 1 divided by Y multiplied by DYDX. In order to solve for the derivative DYDX, we will multiply both sides of the derivative by Y. That will give us D Y D X equal to Y multiplied by the quantity, 3 minus 3 natural logarithm of X, divided by X squared. Now, what, how, what can we do to address this variable of why? Well, earlier, we stated that Y is equal to 3 X, raised the power of 3 divided by X. So we can replace this value of Y with the original function, and we will get D Y D X equal to X, raised to the power of 3 divided by X, multiplied by the quantity, 3 minus 3 natural logarithm of X, divided by X2. If we multiply these equations together, we will further simplify the derivative to be X, raised to the power of 3 divided by X, multiplied by the quantity, 3 minus 3 natural logarithm of X, all divided by X squared. This is going to be the simplest form of our derivative. Now, what we want to do is you want to find the critical points of X in order to find the locations of the local extrema. In order to do this, we want to solve for the values of X that cause the derivative to be zero. Since we have a rational function for a derivative, we will go ahead and solve for the values of X that cause the numerator to be zero. We can take the numerator, which is X, raise the power of 3 divided by X, multiplied by 3 minus 3 natural logarithm of X, and set it equal to 0. By using the zero property, we can set both factors of X equal to 0. We will get X, raise the power of 3 divided by X equal to 0, and 3 minus 3 natural logarithm of X equal to 0. Let's go ahead and solve the first equation. Now, what value of X can we plug in to Xa the power of 3 divided by X, that will cause it to be zero. In order to answer this, we need to think about the domain for a moment. Now, since we have a rational function in the exponent, Since X is located in the denominator, we know that X cannot equal to 0, since that will cause an undefined value. Furthermore, the overall numerator has a natural logarithm of X function in it. That that did not domain of the natural logarithm of X must be that X is strictly greater than 0. What this means is that there are no values of X that we can plug into the first equation that will cause it to equal to 0. So there is no solution to the first factor, thus we will disregard that solution. Now we want to find the values of X that caused 3 minus 3 natural logarithm of X to be zero. We will start by subtracting 3 to both sides of the equation, allowing us to rewrite the equation as -3 natural logarithm of X equal to -3. By dividing both sides of the equation by -3, we can rewrite this equation to be the natural logarithm of X equal to 1. Now we need to remove the X from the natural logarithm. In order to do this, we will exponentiate both sides of the equation. That will leave us with X is equal to E raises the power of 1, which is positive E, and that is going to be the critical point to our derivative. But now we need to determine if this critical point is a local maximum or a local minimum. What we will do is we will go ahead and plot this on the plot this on a number line. So we have the X value E. Now, again, going back to the domain, X must be strictly greater than 0. So what we will do is we will pick testing points to determine the behaviors of the graph between 0 and E and E to positive infinity. For a testing point located to the left of E, we can choose the value X is equal to 1. When X is equal to 1, if we plug this into the original function, we will get the value of 3. Since the function is positive in this region, that means that the graph will be increasing to the left of E. Now we need to pick a testing point to the right of E. Now recall that E is approximately equal to 2.72. So a valid testing point that we can pick to the right of E will be X's equal to 3. When we take the value of 3 and plug it into the function, when F of 3, the value of FF3 is going to equal to approximately negative 0.099. Since the value of the graph is negative to the right of E, the graph will be decreasing to the right of E. And since the graph is increasing. And then decreases when it passes through the value of E. This indicates that E. Is a local maximum. Now, what about the value of the critical point at this value X is equal to E. Well, if we take this and plug it into the function, we get F of E is equal to 3 E, raises the power of 3 divided by E. This is going to be the value of the local maximum. So just to recap, we found the critical points by solving for the derivative and finding the values of X that caused the derivative to be zero. That value came out to be E. We then plotted E on a number line and tested points around the value of E to find the behaviors of the graph. We determined that this had a behavior of being a local maximum, and the value of this local maximum is E raised to the power of 3 divided by E. With that being said, the overall solution to this problem is going to be B. 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.

Local max/min of x¹⸍ˣ Use analytical methods to find all local extrema of the function ƒ(x) = x¹⸍ˣ , for x > 0 . Verify your work using a graphing utility.

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