First of second dirivative test?

Question

Use the first or second dirivative test to find the relative extrema for the following function.

f(x)= x/ lnx

I got most of it done, if someone can just walk me through it so I can check my work. Thanks alot.

Answer 1

Well, f(x)'s domain is (0,∞), so we don't worry about what we see on the graph at (near) (0,0). Let's prove the min we see for 2 ≤ x ≤ 3 is really 3.

f'(x) = [ (ln x)(1) - x(1/x)] / (ln x)² = [ ln x - 1] / (ln x)²

We have a min when f'(x) = 0, so

[ ln x - 1] / (ln x)² = 0
ln x - 1 = 0
ln x = 1
x = e
only the 1 solution.

We evaluate the 2nd derivative at x = e to see if our extremum is a max or min.

f"(x) = 2/[x(ln x)^3] - 1/[x(ln x)²]
f"(e) = 2/[e•1] - 1/[e•1] = 1/e > 0

so our extremum is a minimum, and the only one, since, although the value of f(x) < 0 for 0 ≤ x ≤ 1, x=1 is an asymptote, so f(x) is undefined for x=0 and x=1.

Answer 2

For the derivative you get
f'(x) = (ln x - 1)/(ln x)^2
Set the derivative = 0 and solve for x. You get
ln x = 1, or
x = e.

Then notice that
f'(x) < 0 if 0 < x < e and
f'(x) > 0 if e < x.

So the graph is decreasing to the left of (e,f(e)) and increasing to the right of this point. That makes (e,f(e)) a global minimum (or absolute minimum).

Answer 3

y'=(lnx-1)/(lnx)^2
y'=0<--->lnx=1<-->x=e
For x (0,e)<--->f'(x)<0
For x (e,inf)<--->f'(x)>0
For x=e you have local
minimum y=e
You cannot aplly sec.der.test
because y''=0 for x=e