Question

f(x)=x^(1/x) how do i find the max value using calculus?

Answer 1

1. We say that f(x) has an absolute (or global) maximum at if for every x in the domain we are working on.
2. We say that f(x) has a relative (or local) maximum at if for every x in some open interval around .
3. We say that f(x) has an absolute (or global) minimum at if for every x in the domain we are working on.
4. We say that f(x) has a relative (or local) minimum at if for every x in some open interval around .

Answer 2

its really easy hope that helps

Answer 3

no not really, i need to find out how to get a number out of it.

Answer 4

@ArmyBoy97 you can't just copy and paste and not know what it means.

Answer 5

thanks

Answer 6

1/ take derivative
2/ let it =0 to find x
3/ plug back to the equation to find min/max

Answer 7

thanks let me try it

Answer 8

is the derivative 1/x(x)(lnx)-0

Answer 9

=0
i mean

Answer 10

@OOOPS

Answer 11

Let y = x^(1/x)
ln both sides
lny = 1/x lnx
derivative both sides
\[\dfrac{y'}{y}= (\dfrac{1}{x})'lnx+(lnx)'\dfrac{1}{x}\\y'=-y(\dfrac{lnx}{x^2}+\dfrac{1}{x^2})\]
replace y to have \(y'=-x^{1/x}(lnx+1)/x^2\)
this expression =0 iff lnx+1 =0 --> lnx =-1 --> x = 1/e
that is the value of x which make the graph min or max. However, you have just 1 value of x, you can't define whether it is min or max
You must take second derivative to consider whether the concave is up or down, if it is up--> this point is min; if it is down, --> this point is max.
I mean:

Answer 12

\[y=(x ^{1/2}(1-\ln(x))/x^2\]

Answer 13

why x^1/2?? x^(1/x)

Answer 14

oh oops
k let me try again

Answer 15

is it supposed to be y(\[y(1/x ^{2}-lnx/x ^{2})\]

Answer 16

Oh yeah, you are right. My bad. hihihi...

Answer 17

coool i was like damn what did i do wrong

Answer 18

can you show me how to do the second derivative to

Answer 19

From y'= y(1/x^2-lnx/x^2)
y" = y'(1/x^2-lnx/x^2) + y(1/x^2-lnx/x^2)'
replace y' and y after take derivative the last term