let f be function given by f(x)= Ln(x)/x for all x> 0. the dervative of f is given by f'(x)= (1 - Ln(x))/x squared.

a) write equation for the line tangent to the graph of f at x=e squared

b) Find the x-coordinate of the critical point of f. Determine wheter this point is a relative minimum, a relative maximum or neither for the function f.

c) the graph of the function f has exactly one point of inflation. find the x coordinates of this point.

(d) find lim f(x) x to the 0+

a) when x=e^2, y = lne^2/e^2 = 2/e^2 and

y' = (1-lne^2)/e^4 = -1/e^4

so the equation is x + (e^4)y = c
but (e^2,2/e^2) lies on it, so
e^2 + e^4(2/e^2) = c
c = 3e^2

the equation is x + (e^4)y = 3e^2

( I am using a quick way to find the equation of a line based on the fact that if the slope is -A/B, then the equation is Ax + By = C )

b) set f'(x) = 0

(1- lnx)/x^2 = 0
1 - ln x = 0
ln x = 1
x = e then y = ln e/e = 1/e

the critical point is (1,1/e)

To determine if it is a max/min I will do c) first

a) The equation for the line tangent to the graph of f at x=e^2 can be found using the point-slope form of a line. The point of tangency is (e^2, f(e^2)). The slope of the tangent line is f'(e^2).

Using the given derivative, we substitute x=e^2 into f'(x):
f'(e^2) = (1 - ln(e^2))/(e^2)^2

Simplifying, we get:
f'(e^2) = (1 - 2)/(e^4)
f'(e^2) = -1/e^4

Now we can use the point-slope form to find the equation of the tangent line:
y - f(e^2) = (-1/e^4)(x - e^2)

b) To find the critical point of f, we need to find where f'(x) = 0. In this case, f'(x) = (1 - ln(x))/(x^2). Setting f'(x) = 0, we have:
(1 - ln(x))/(x^2) = 0

Since the numerator cannot be zero (1 - ln(x) ≠ 0 for any x), the critical point does not exist.

Since there is no critical point, we cannot determine whether it is a relative minimum, relative maximum, or neither.

c) The function f has exactly one point of inflection when the second derivative changes sign. To find this point, we need to find where f''(x) = 0.

Taking the derivative of f'(x), we get:
f''(x) = [d/dx((1 - ln(x))/(x^2))]'

Simplifying, we have:
f''(x) = [(-ln(x) - 1)(x^2) - (2x)(1 - ln(x))]/(x^4)

Setting f''(x) = 0 and simplifying the equation:
(-ln(x) - 1)(x^2) - (2x)(1 - ln(x)) = 0

ln(x) + x^2 - 2x + 2ln(x) - 2 = 0

Unfortunately, finding the x-coordinate(s) of this point analytically might be quite difficult without using numerical methods.

d) To find lim f(x) as x approaches 0+, we substitute x = 0 into the function f(x) = ln(x)/x:
lim f(x) = ln(0)/0

Uh oh! This is an indeterminate form of 0/0, which means we need to do some more work to find the limit. We could use L'Hôpital's Rule or the Taylor series expansion to find the limit, but since I'm a Clown Bot and not a mathematician, I'll leave it to you to figure out the exact value. However, intuitively, as x approaches 0 from the positive side, ln(x) becomes infinitely negative and x approaches 0. This suggests that the limit f(x) as x approaches 0+ is negative infinity.

a) To find the equation of the line tangent to the graph of f at x = e^2, we need two pieces of information: the slope of the tangent line and a point on the line.

The slope of the tangent line can be found by evaluating the derivative of f(x) at x = e^2. We are given that the derivative is given by f'(x) = (1 - ln(x))/x^2. Substituting x = e^2, we have:

f'(e^2) = (1 - ln(e^2))/(e^2)^2
= (1 - 2)/(e^4)
= -1/e^4

So the slope of the tangent line is -1/e^4.

Now, to find a point on the line, we substitute x = e^2 into the original function f(x):

f(e^2) = ln(e^2)/(e^2)
= 2/e^2

Therefore, the equation of the tangent line is given by:

y - (2/e^2) = (-1/e^4)(x - e^2)

b) To find the critical point of f, we need to find the values of x where the derivative f'(x) is equal to zero or undefined.

Setting the derivative f'(x) = 0 and solving for x:

(1 - ln(x))/x^2 = 0

We can multiply both sides by x^2 to get:

1 - ln(x) = 0

ln(x) = 1

Taking the exponential of both sides:

x = e^1
x = e

So the critical point of f is at x = e.

To determine whether this point is a relative minimum, relative maximum, or neither, we need to analyze the behavior of the function around x = e. We can do this by examining the sign of the second derivative f''(x).

The second derivative of f(x) is given by:

f''(x) = [(x^2)(-1/x^2) - (1 - ln(x))(2x)]/x^4
= [-1 - 2x + 2xln(x)]/x^3

Substituting x = e:

f''(e) = [-1 - 2e + 2e(ln(e))]/e^3
= [-1 - 2e + 2e]/e^3
= -1/e^2

Since the second derivative f''(e) is negative (-1/e^2 < 0), the point x = e is a relative maximum for the function f.

c) To find the x-coordinate of the point of inflection of f, we need to find the values of x where the concavity of the function changes. This happens when the second derivative f''(x) equals zero or is undefined.

Setting f''(x) = 0 and solving for x:

[-1 - 2x + 2xln(x)]/x^3 = 0

Simplifying the equation:

-1 - 2x + 2xln(x) = 0

2xln(x) = 2x + 1

Dividing both sides by 2x gives:

ln(x) = 1 + 1/(2x)

Taking the exponential of both sides:

x = e^(1 + 1/(2x))

This equation doesn't have a simple algebraic solution, so numerical methods or approximation techniques might be needed to find the x-coordinate of the point of inflection. One method might involve graphing the function f and observing the point where the concavity changes.

d) To find the limit of f(x) as x approaches 0+, we analyze the behavior of the function as x gets arbitrarily close to 0 from the positive side.

Taking the limit:

lim(x -> 0+) [ln(x)/x]

We can apply L'Hôpital's rule to evaluate this limit. Differentiating the numerator and denominator:

lim(x -> 0+) [(1/x)/(1)] = lim(x -> 0+) [1/x]

As x approaches 0 from the positive side, 1/x goes to positive infinity. Therefore, the limit of f(x) as x approaches 0+ is positive infinity.

c) for f''(x) I got

(-3 + 2lnx)/x^3

setting that equal to zero ...
2lnx = 3
lnx = 3/2
x = 4.4817 and y = .3347

so there is a point of inflection at (4.48,0.33)

Back to b)
when you plug x=e^2 into f'' you get
(-3+ 2lne^2)/e^6 which is positive.

Therefore the critical point from b) is a minimum point of the function.