Here's an alternative way to solve: Exponentiating the function: f(x) = x^(1/x) = e^[(1/x) * ln(x)] Derivative: f'(x) = e^[(1/x) * ln(x)] * [(-1/x^2) * ln(x) + (1/x^2)] We set f'(x) to 0 to find critical points: 0 = e^[(1/x) * ln(x)] * [(-1/x^2) * ln(x) + (1/x^2)] Ignore "e^[(1/x) * ln(x)]" as it never equals 0 0 = (-1/x^2) * ln(x) + (1/x^2) -(1/x^2) = (-1/x^2) * ln(x) Cancel out "-(1/x^2)" from both sides: 1 = ln(x) Logarithmic to exponential form: x = e^1 x = e There is a critical point at x = e. Subsitute e into f(x) to get: f(e) = e^(1/e) Using the second derivative (which is very long so I'm going to skip showing it), plugging in "e" gives us a negative number, meaning it is concave down (i.e., local maximum). Thus, we can conclude that (e, e^(1/e)) is the maximum point of the function.
Hey man, I really loved the video. Always had trouble with this subject and I thought I was dumb Turns out I just didn't have a good teacher The way you explain things were precise and understandable Keep up the good work
Bro didn't tell us what the limit as x goes to infinite of x^(1/x) is but just carefully dodged it by saying we aren't talking about it. Obviously it can't be evaluated at infinity but the limit which is interesting can be. Oh yea in case anyone wonders, it converges and results with 1...
fun fact: x^(1/x) for 1/e < x < e is the inverse function for infinite tetration.
Here's an alternative way to solve:
Exponentiating the function:
f(x) = x^(1/x) = e^[(1/x) * ln(x)]
Derivative:
f'(x) = e^[(1/x) * ln(x)] * [(-1/x^2) * ln(x) + (1/x^2)]
We set f'(x) to 0 to find critical points:
0 = e^[(1/x) * ln(x)] * [(-1/x^2) * ln(x) + (1/x^2)]
Ignore "e^[(1/x) * ln(x)]" as it never equals 0
0 = (-1/x^2) * ln(x) + (1/x^2)
-(1/x^2) = (-1/x^2) * ln(x)
Cancel out "-(1/x^2)" from both sides:
1 = ln(x)
Logarithmic to exponential form:
x = e^1
x = e
There is a critical point at x = e.
Subsitute e into f(x) to get:
f(e) = e^(1/e)
Using the second derivative (which is very long so I'm going to skip showing it), plugging in "e" gives us a negative number, meaning it is concave down (i.e., local maximum).
Thus, we can conclude that (e, e^(1/e)) is the maximum point of the function.
this is actually completely expected
Great video!
nice video! keep it up!
Hey man, I really loved the video.
Always had trouble with this subject and I thought I was dumb
Turns out I just didn't have a good teacher
The way you explain things were precise and understandable
Keep up the good work
Crazy how the intro feels like you're about to start playing minecraft....
with questions like these, as a person who is not that good at math i just assume the answer is e
Was very surprised to end this video and see that you have less than 50 subscribers ngl, thought it would be way more
Shorcut could be substituting y'=0 immediately at 4:51 and simplifying from there.
i like your funny words magic man
nice video mate
why isn't this channel more popular??
Bro didn't tell us what the limit as x goes to infinite of x^(1/x) is but just carefully dodged it by saying we aren't talking about it. Obviously it can't be evaluated at infinity but the limit which is interesting can be. Oh yea in case anyone wonders, it converges and results with 1...