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f(x)=(lnx)/(5sqrt(x))a. Find the interval on which f is increasing and decreasing....

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rachelgree | Student, College Freshman | eNoter

Posted April 8, 2012 at 12:52 AM via web

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f(x)=(lnx)/(5sqrt(x))

a. Find the interval on which f is increasing and decreasing.

b. Find the local maximum value of f

c. Find the inflection point

d.Find the interval on which f is concave up and down

2 Answers | Add Yours

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 8, 2012 at 8:27 AM (Answer #1)

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You should remember that  the first derivative of function and the critical values of function tell you about monotony of function, hence you need to differentiate the function with respect to x such that:

f'(x) = ((lnx)'*(5sqrt(x)) - (lnx)*(5sqrt(x))')/(25x)

f'(x) = ((1/x)*(5sqrt(x)) - (lnx)*(5/2sqrt(x)))/(25x)

You need to solve the equation f'(x) = 0 such that:

5sqrtx/x - 5lnx/(2sqrtx) = 0

You should bring the terms to a common denominator such that:

10sqrtx - 5lnx*sqrtx = 0

You need to factor out 5sqrtx such that:

5sqrtx(2 - ln x) = 0

You should solve the equations 5sqrtx=0 and 2-ln x=0 such that:

5sqrtx = 0 => x = 0

2 - ln x = 0 => ln x = 2 => x = e^2

Notice that for the value x=0 , the equation ((1/x)*(5sqrt(x)) - (lnx)*(5/2sqrt(x)))/(25x) does not hold, hence you should keep only x = e^2.

You need to select a value for x, smaller than x=e^2 such that:

x = e => 5sqrt e(2 - ln e) = 5sqrt e > 0

You need to select a value for x, larger than x=e^2 such that:

x = e^3 => 5sqrt (e^3)(2 - 3) = -5sqrt e < 0

Hence, the derivative is negative for x in (e^2,oo), hnece the function decreases over the interval and derivative is positive for x in (0,e^2), hence the function increases over interval.

b) Notice that the critical value of the function is x=e^2 and the evaluation of the monotony of function yileds that the function reaches its maximum at x = e^2 => f(e^2) = (lne^2)/(5sqrt(e^2))

f(e^2) = 2/(5e)

c) You need to find the roots of second derivative to find the inflection point of function such that:

f''(x) = ((5sqrtx/x - 5lnx/(2sqrtx))'*(25x) - (5sqrtx/x - 5lnx/(2sqrtx))*(25x)')/(625x^2)

f''(x) = (((5x/(2sqrtx) - 5sqrtx)/x^2 - (10sqrtx/x - 5lnx/sqrtx)/(4x))*(25x) - 25(5sqrtx/x - 5lnx/(2sqrtx)))/(625x^2)

f''(x) = (((10sqrtx - 20sqrtx - 10sqrtx + 5lnx*sqrtx)/(4x^2))*(25x) - 25(5sqrtx/x - 5lnx/(2sqrtx)))/(625x^2)

f''(x) = ((( - 20sqrtx + 5lnx*sqrtx)/(4x)) - (5sqrtx/x - 5lnx/(2sqrtx)))/(25x^2)

You should solve the equation f''(x) = 0 => ((( - 20sqrtx + 5lnx*sqrtx)/(4x)) - (5sqrtx/x - 5lnx/(2sqrtx)))=0

-5sqrtx/x + (5lnx*sqrtx)/(4x) -  5sqrtx/x + 5lnx/(2sqrtx) = 0

-2sqrtx/x + (lnx*sqrtx)/(4x) + lnx/(2sqrtx) = 0

-8sqrtx + lnx*sqrtx + 2sqrtx*lnx = 0

-8sqrtx + 3lnx*sqrtx = 0

You need to factor out sqrtx such that:

sqrtx(-8 + 3lnx) = 0 => sqrtx = 0 => x = 0

3ln x = 8 => ln x = 8/3 => x = root(3)(e^8) => x = e^2*root(3)(e^2)

Hence, since x = 0 is excluded as root, then the inflection point of the function has x coordinate at x = e^2*root(3)(e^2).

d) You need to select a value for x larger than x = e^2*root(3)(e^2) such that root(3)(e^9) = e^3 such that:

sqrt (e^3)(-8 + 9) = sqrt (e^3) > 0

You need to select a value for x smaller than x = e^2*root(3)(e^2) such that root(3)(e^6) = e^2 such that:

sqrt (e^2)(-8 + 6) = -2e < 0

Hence, the graph of function is concave up over (e^2*root(3)(e^2), oo ) and it is concave down over (0, e^2*root(3)(e^2)).

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 8, 2012 at 8:31 AM (Answer #2)

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a)You should remember that the first derivative of function and the critical values of function tell you about monotony of function, hence you need to differentiate the function with respect to x such that:

`f'(x) = ((lnx)'*(5sqrt(x)) - (lnx)*(5sqrt(x))')/(25x)`

`f'(x) = ((1/x)*(5sqrt(x)) - (lnx)*(5/2sqrt(x)))/(25x)`

You need to solve the equation f'(x) = 0 such that:

`5sqrtx/x - 5lnx/(2sqrtx) = 0`

You should bring the terms to a common denominator such that:

`10sqrtx - 5lnx*sqrtx`  = 0

You need to factor out `5sqrtx`  such that:

`5sqrtx(2 - ln x) = 0`

You should solve the equations `5sqrtx=` 0 and `2-ln x=0`  such that:

`5sqrtx = 0`  => x = 0

`2 - ln x = 0 =gt ln x = 2 =gt x = e^2`

Notice that for the value x=0 , the equation `((1/x)*(5sqrt(x)) - (lnx)*(5/2sqrt(x)))/(25x)`  = 0 does not hold, hence you should keep only `x = e^2` .

You need to select a value for x, smaller than `x=e^2`  such that:

`x = e =gt 5sqrt e(2 - ln e) = 5sqrt e gt 0`

You need to select a value for x, larger than `x=e^2`  such that:

`x = e^3 =gt 5sqrt (e^3)(2 - 3) = -5sqrt e lt 0`

Hence, the derivative is negative for `x in (e^2,oo),`  hnece the function decreases over the interval and derivative is positive for`x in (0,e^2), ` hence the function increases over interval.

b) Notice that the critical value of the function is `x=e^2`  and the evaluation of the monotony of function yields that the function reaches its maximum at  `x = e^2 =gt f(e^2) = (lne^2)/(5sqrt(e^2))`

`f(e^2) = 2/(5e)`

c) You need to find the roots of second derivative to find the inflection point of function such that:

`f''(x) = ((5sqrtx/x - 5lnx/(2sqrtx))'*(25x) - (5sqrtx/x - 5lnx/(2sqrtx))*(25x)')/(625x^2)`

`f''(x) = (((5x/(2sqrtx) - 5sqrtx)/x^2 - (10sqrtx/x - 5lnx/sqrtx)/(4x))*(25x) - 25(5sqrtx/x - 5lnx/(2sqrtx)))/(625x^2)`

`f''(x) = (((10sqrtx - 20sqrtx - 10sqrtx + 5lnx*sqrtx)/(4x^2))*(25x) - 25(5sqrtx/x - 5lnx/(2sqrtx)))/(625x^2)`

`f''(x) = ((( - 20sqrtx + 5lnx*sqrtx)/(4x)) - (5sqrtx/x - 5lnx/(2sqrtx)))/(25x^2)`

You should solve the equation `f''(x) = 0 =gt ((( - 20sqrtx + 5lnx*sqrtx)/(4x)) - (5sqrtx/x - 5lnx/(2sqrtx)))=0`

`-5sqrtx/x + (5lnx*sqrtx)/(4x) - 5sqrtx/x + 5lnx/(2sqrtx) = 0`

`-2sqrtx/x + (lnx*sqrtx)/(4x) + lnx/(2sqrtx) = 0`

`-8sqrtx + lnx*sqrtx + 2sqrtx*lnx = 0`

`-8sqrtx + 3lnx*sqrtx = 0`

You need to factor out `sqrtx`  such that:

`sqrtx(-8 + 3lnx) = 0 =gt sqrtx = 0 =gt x = 0`

`3ln x = 8 =gt ln x = 8/3 =gt x = root(3)(e^8) =gt x = e^2*root(3)(e^2)`

Hence, since x = 0 is excluded as root, then the inflection point of the function has x coordinate at `x = e^2*root(3)(e^2).`

d) You need to select a value for x larger than `x = e^2*root(3)(e^2)`  such that `root(3)(e^9) = e^3`  such that:

`sqrt (e^3)(-8 + 9) = sqrt (e^3) gt 0`

You need to select a value for x smaller than `x = e^2*root(3)(e^2) ` such that `root(3)(e^6) = e^2`  such that:

`sqrt (e^2)(-8 + 6) = -2e lt 0`

Hence, the graph of function is concave up over `(e^2*root(3)(e^2), oo )`  and it is concave down over `(0, e^2*root(3)(e^2)).`

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