finding the absolute minimum and maximum of f(x)=x^-2*ln(x)

[@田村博志-z8y]

We have to show that 1/2 < √e < 4. This is very simple. For every real numbers x satisfying 0 < x < 2, we have 1 + x < e^x < ( 2 + x )/( 2 - x ). Applying x = 1/2, we have 3/2 < √e < 5/3. Then we obtain 1/2 < √e < 4. Of course, if the fact that 2.7 < e < 2.8 is given, then it is clear that 1/2 < √e < 4. But it is important to check that f has a critical point in its domain.

[@gennaroponsiglione1098]

How did you find the inequality e^x

[@田村博志-z8y]

@@gennaroponsiglione1098 Let a, b be real numbers satisfying a < b. Let f be a continuous function on [ a, b ]. We consider the approximation of the integral ∫_{ a }^{ b } f( t ) dt. Assume that f is convex. Then we have ∫_{ a }^{ b } f( t ) dt < ( 1/2 )( b - a )( f( a ) + f( b ) )　…① where the right hand side of ① is the area of a trapezoid. Let x > 0. Applying f( t ) := e^t, a := 0, b := x to ①, we have ∫_{ 0 }^{ x } e^t dt < ( 1/2 )・x・( 1 + e^x ), e^x - 1 < ( x/2 )( 1 + e^x ), 2・e^x - 2 < x + x・e^x, ( 2 - x )・e^x < 2 + x.　…② If x >= 2, then ② is trivial and useless. So we assume that 0 < x < 2. Then we obtain e^x < ( 2 + x )/( 2 - x ).

[@gennaroponsiglione1098]

@@田村博志-z8y very nice, thanks

[@YorangeJuice]

this is not really related to this video, but can u do more calculus 3 content plz

[@melissasik2483]

thank you!!

[@pixelpix1728]

Question, why did you consider trying to set the denominator to 0? That doesn't approach 0, that approaches infinity, it's the opposite of what we were looking for, no?

[@ΛυμπέρηςΧασάπης]

Critical numbers of a function are 1. The roots of the derivative 2. The ones that make the derivative undefined (x is in the domain of f(x) but not in the domain of f'(x)) He was looking for critical numbers so he considered both cases

[@stephenbeck7222]

If the derivative approaches infinity, that means you either have a vertical asymptote or a cusp on the function. The vertical asymptote won’t be an absolute extremum of the function (actually it would cause there to not be an abs min and/or abs max) but a cusp could be an absolute extremum.

[@risheraghavendira1932]

Please heart me @just calculus