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Numerade Educator



Problem 52 Medium Difficulty

Use the guidelines of this section to sketch the curve.

$ y = \dfrac{\ln x}{x^2} $


See attached photo.


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Video Transcript

So the domain for this function is X. It has to be greater than zero. Um, because we can't have a zero because of the denominator. And we can't have negative numbers because of the numerator intercepts. 10 symmetry. None. Essen totes. We have one X equals zero again because the bottom cannot equal zero. And we also have ah, horizontal one. Because the limit as X approaches infinity, um, of our function. Ellen of X over X squared is zero. So we have, um as X approaches infinity, we have ah, horizontal one at y equals zero. Okay. And next we confined the first derivative. So why praying? My prime equals one minus two ln x over X cute set that equal to zero. And we find, um the ex equals the square root of E, which is approximately 1.6. Um, so we have a max at 1.6 because the first derivative test tells us that, um, 1.6. So values below 1.6 will be increasing above will be decreasing if you test it in the first derivative. So, um, 1.6 is a max. So And by the way, the point is 1.6 comma, 0.18 if we substitute, um into our original function. So 1.60 point 18 that's her first derivative test. And now we confined our second derivative, Furqan Cavaney. So our second derivative is why, um, double prime equals Negative. Negative. Six Ellen X plus five over X to the power of four. Set that equal to zero. Um, in Seoul for ex um, so you can see that X equals e to the five over six, which is approximately 2.3. Okay. And, um, f of 2.3 is zero. Oops. Up of 2.3 is about 0.15 zero point. So our second derivative test tells us that 2.3. If we test values, um, into the second derivative that are smaller, we'll see it's conk it down. So Colin Cave don't. And then Khan gave up. If it's above 2.3. So this is an inflection point at 2.3 comma, 1.0 point 15. Okay, so that's all the information we have. And we can graft now. Okay. So we can graph our, um, Assam totes. So we have when X equals zero. So that's our vertical one. And we also have one at why equals zero. That's a horizontal one. So we have ah intercept at 10 So let's say this is one 10 and we have, um, a Max at 1.6. So that's about here. Um, in 0.18. So this is our max. This is just an approximate It's the sketch, Um, and we have It's gonna be increasing until it hits 1.6. So it's gonna be increasing. Go through the, um, intercept. And once it has 1.6, it's going to start decreasing. So now it's gonna go down, but it's still gonna be con cave up until we hit 2.3. So let's say, um, if this is one, let's say that so 012 this is about two. So 2.3 would be here, So it's gonna, um although it's gonna decrease its going to still be con cave down until about here, where it switches to con cave up. And although it's conclave down, it's gonna continue decreasing and approaching but never touching our ass in tow.