Question

SOMEONE HELP ME D:!!! What value of x does f(x)=x^2-lnx have an absolute minimum?

Answer 1

Take the derivative and then set the differential equation to zero. That should mark either a maximum or a minimum. What depends is the concavity of the original function.

Answer 2

f'(x) = 2x - 1/x = 0
2x^2 - 1 = 0
x = +- sqrt(1/2)

Answer 3

the negative value of x is inadmissable because a negative log is undefines
soe the only turning point is at x = sqrt(1/2)
second derivative is 2 + 1/x^2 which is postive for x= sqrt1/2 so
this gives a minimum

so x^2 - ln x has a minimum value when x = sqrt(1/2)
if u go to wolframalpha.com and enter the fuction it will show you the graph and u'll see it is an absolute minimum