f(x)=lnx/x for x>0?

Question

f(x)=lnx/x for x>0, why is this statement true

f is decreasing for all x is greater than e.
can you explain where e came from and just everything because i don't get this question at all

Answer 1

f(x) = ln(x)/x

To determine the intervals of increase/decrease, find the first derivative and make it 0.

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

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

Our critical values are what make f'(x) = 0 or what make
f'(x) undefined. Equating the numerator to 0 makes it 0, and equating the denominator makes it undefined.

In our case, our critical values are found by solving

1 - ln(x) = 0, and x^2 = 0

1 - ln(x) = 0 means
ln(x) = 1, and converting this to exponential form
x = e^1 = e

x^2 = 0 means x = 0.

Critical values: 0, e

Make a number line.

. . . . . . . . . . . . . (0) . . . . . . . . . . . (e) . . . . . . . . . . .

And test a single value (for f'(x) in each region for positivity and negativity.

We don't have to worry about the region less than 0 because the domain of ln(x) prevents any values from being negative.

In the 2nd region, test x = 1. Then
f'(1) = [1 - ln(1)]/1^2 = [1 - 0]/1 = 1, which is positive. Mark the region as positive.

. . . . . . . . . . . . . (0) . . . . . {+}. . . . . (e) . . . . . . . . . . .

Test a value greater than e. The useful thing is that we can test ANY single value greater than e, no matter how big. For that region, let's test 100000. Then

f'(100000) = [1 - ln(100000)]/100000^2

We get a negative number over a a positive number, which is negative. Mark the region as negative.

. . . . . . . . . . . . . (0) . . . . . {+}. . . . . (e) . . . . . {-} . . . .

Our function is increasing in the positive regions, and decreasing in the negative regions.

f(x) is increasing on (0, e]

f(x) is decreasing on [e, infinity)

Answer 2

Ok, you have to analyze the F(x) :

F(x) = lnx / x

x cannot be cero, because, if it's cero, F(0) = not determinate

And if you analyze : lnx, you know that x can only be more than cero : x > 0, so the domain will be x > 0

If you wanna graphic the F(x) :

dF(x) / dx = -lnx*x^-2 + 1 = 0

lnx*x^-2 -1 = 0

So "e" is coming up when you analize the critical points of the graphic :

lnx*x^-2 = 1

x^2 = lnx >>> e^(x^2) = x

And yes, the graphic will be decreasing for all x greater than e.

Take "e" as a critical point, and then analyze the points :

x < e and x > e, in :

F'(x) = lnx*x^-2 -1

Hope that might help you

Answer 3

If you differentiate this function you find that :
d(f(x))/dx = 1/(x^2) - lnx/(x^2).
We have to solve the equation d(f(x))/dx = 0 , that is:
lnx = 1 which implies x = e.
For x>e we have lnx >1 and so d(f(x))/dx < 0 which means that f is decreasing for x>0.

Answer 4

1) I'm not sure I get the question either; but, consider the following comments

2) If it said f(x) = ln(x) / x; for all x, then the statement would be wrong because ln(x) is undefined for x <= 0. Try it on your calculator.

3) So, the statement simply says that f(x) = ln(x) / x is true for all x > 0

4) Note that f(x) approaches 0 as x approaches infinity

Answer 5

calculate f'(x) =1/x^2 *( 1-lnx)
1-ln x= 0 ln x=1 and x=e
Sign of f' before x=e it is positive so f increase .After x=e f' is negative so f decrease. f(e) is a relative maximum

Answer 6

well.. e it's the base of the natural logharitm. e it's about 2.71..(if i remember well)... and if X it's greater than e than lnx it's greater than 1. and since X it's greater than 1 that function it's decreasing... it goes to 0 to be more precise.. i mean its limit when x goes to infinite it's 0.
cheers !

Answer 7

if x was less than zero, you would get a non-real answer making the statement false. And e is a number. Ln is really log base e.

Answer 8

To find whether a function is increasing or decreasing you should differentiate it. If the derivative is positive it is increasing, if negative it is decreasing.

Answer 9

f'(x) = [x*d(ln x)/dx - lnx] /x^2
=[ x*1/x - ln x]/x^2
=[ 1 - ln x]/x^2
for f(x) to be decreasing f'(x)<0
or, [ 1 - ln x]/x^2 < 0
or, 1 - ln x<0
or, 1< ln x
or, e^1 or, x >e
so for all x>e, f(x) is decreasing

Answer 10

answer is

E = MC HAMMER