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Numerade Educator

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Problem 44 Hard Difficulty

Find the critical numbers of the function.

$ f(x) = x^{-2} \ln x $

Answer

The only critical number of the given function is $x=\sqrt{e}$

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Video Transcript

we want to find the critical numbers of the function F of X equals X to the negative two times natural logarithms of X. So we remember the definition of critical number. We say that a number C. In the domain of the function F. It's a critical number of the function. If either the first derivative at that number is zero or first derivative at that number does not exist. So this is the definition of critical number. And then we calculate the derivative. But first we remember here, which is the domain of these functions as we can see, we have this is the same as natural logarithms of X over eggs square. So we have a fraction division of functions. So it's important that the denominator is not zero. So X cannot be zero. But also the natural algorithm imposed the condition and eggs to be positive. And in fact if X is positive, it is already different from zero. And so the general condition of the points X to evil they currently dysfunction is that X. B. Positive. So we can say that the main is The Interval zero Plus Infinity. That is the real numbers. So she's dad eggs is positive, that's the main. And is there in this interval where we look at the curriculum numbers by definition are points or numbers into the main of the function. So now we calculate the first derivative of F. We have a product, we're going to use this form instead of this product here. So we apply the proper role of the derivative and we get derivative extra native to is native to X To the negative three times natural rhythm of X Plus X to the -2 times. These are derivative of natural algorithm effects. She's one of rex and this is the same as negative, negative to natural rhythm of X over X cube plus. Then he's an anti particle putting the denominator, we get one over X cube. So this is the same as 1 -2. natural rhythm of eggs over eggs Q. That is the first relative is equal to one minus two natural rhythm of X over eggs Q. And so let's see if there are points in the domain where conservative is not defined. In fact we have again, natural algorithm effects which is a fine for all positive numbers. And the division the fraction is possible When X is not zero. So two conditions again exist. Material X is positive. Give us that X must be positive, which includes already that X is not zero. So uh these relativity fine exist, let's say better exist or every point or value eggs into the main of F. That is for every positive value eggs. And that means that we won't find critical points with the condition that the derivatives that does not exist. So the only critical point or critical numbers of the given function F are those values those X. Positive in the real numbers, search that the first derivative E. Zero. So we got to solve the equation after the difficult zero for positive values of X. So we start with first serve a difficult zero. It's the same saying that 1 -2 natural rhythm of x over X. Q. is zero. And this is the same as saying that 1 -2 natural rhythm effects zero. And that's important to remark that this equivalence implies that we are already uh the values of X that are positive. That is any of these two sides of the equivalence imposed this condition because we have the natural liberation in both of them. And so we have this uh equation here which is the same as Natural sodium of eggs equal to 1/2. And that's the same as E. To the natural rhythm of X equal to E. To the 1/2 power. And that's the same as remember the exponential and the natural logarithms are in their functions of each other. So that gets two eggs Equal Italy 1/2 power in that Square TV. That is the only solution of this equation is a positive number X equal to scores of E. So the only critical number after given function F. Is X equal squirrel of E. So it's very important that we uh find first domain of the function they ended the find the derivative. Look at where the a derivative exist if there are points in domain where derivative does not exist. They are critical points or numbers. And besides that we got to find where the derivative is equal to zero. So we have done that and we found only one critical point in this example