(a) find the intervals on which f is incrs or decrs.
(b) find the local min/max values of f.
(c) find the intervals of concavity and the inflection points.
f(x)=(lnx)/sqrtx
I took the derivative and got
2-lnx/2x^3/2
take the values when f'(x) >0 I get
2-lnx > 0
2 > lnx
now how do get x by itself on one side?
(b) how do I got about this?
(c) take f''(x, set numerator equal to zero and solve but how?
you should use brackets to obtain the proper order of operation,
f '(x) = (2- lnx)/(2x^(3/2))
why don't we do part b) first, then
2 - lnx = 0
lnx = 2 , by definition lnx = 2 <----> e^2 = x
so e^2 = x = appr. 7.39
so when x=e^2 we have a max/min value of the function
f(e^2) = ln(e^2)/√(e^2) = 2/2.718 = appr. .736
okay I get it now! But how do I find (C)find the intervals of concavity and the inflection points.
I think I would take the 2nd derivative and set the numerator equal to zero and solve but i'm not quite sure how solve it.
To find the intervals on which f is increasing or decreasing (part a), you correctly took the derivative of f(x) and obtained:
f'(x) = (2 - ln(x)) / (2x^(3/2))
To determine where f'(x) > 0, you set the numerator (2 - ln(x)) greater than zero:
2 - ln(x) > 0
To isolate ln(x), you can continue solving as follows:
-ln(x) > -2 (subtract 2 from both sides)
ln(x) < 2 (negate both sides, which reverses the inequality)
Now, to get x by itself, you can exponentiate both sides with the base e (natural logarithm base):
e^(ln(x)) < e^2
x < e^2
Hence, the solution for x would be any value less than e^2. You can write this interval as (-∞, e^2).
For finding the local minimum and maximum values of f (part b), you need to locate the critical points where f'(x) equals zero or is undefined. In this case, there is only one critical point, and it occurs when the numerator is zero:
2 - ln(x) = 0
To solve for x, you can manipulate the equation as follows:
-ln(x) = -2
ln(x) = 2
Again, exponentiating both sides:
e^(ln(x)) = e^2
x = e^2
So the critical point is x = e^2.
To determine if this point is a local minimum or maximum, you need to analyze the concavity of f. This leads us to part c.
To find the intervals of concavity and the inflection points (part c), you need to take the second derivative of f(x). You previously found the first derivative as:
f'(x) = (2 - ln(x)) / (2x^(3/2))
Now, differentiate f'(x) with respect to x:
f''(x) = (-1/x) / (2x^(3/2)) = -1 / (2x^(5/2))
To find where f''(x) = 0, you set the numerator equal to zero:
-1 = 0
Since -1 is never equal to zero, there are no values of x where f''(x) equals zero. Therefore, there are no inflection points.
To determine the intervals of concavity, evaluate f''(x) at a test point within each interval found in part a. The intervals where f''(x) is positive correspond to concave up, and the intervals where f''(x) is negative correspond to concave down.
Let's test two points: x = 1 and x = 4. Plug these values into f''(x):
f''(1) = -1 / (2 * 1^(5/2)) = -1/2 (< 0)
f''(4) = -1 / (2 * 4^(5/2)) = -1/32 (< 0)
Both values are negative, indicating that f(x) is concave downward over the entire domain.
Thus, the final answers for the given function f(x) = (ln(x))/sqrt(x) are:
(a) f is decreasing on the interval (-∞, e^2) and increasing elsewhere.
(b) The only critical point is x = e^2, and it needs further analysis to determine if it is a local minimum or maximum.
(c) f(x) is concave downward for all values of x, and there are no inflection points.