How do you find increasing, decreasing, inflection points, minimum and maximum for the graph #f(x) = ln(x)/(8sqrtx)#?

Answer 1

See the explanation.

The first and second derivatives must be found:

#f'(x)=(1/x*8sqrtx-8/(2sqrtx)lnx)/(64x)=(1/sqrtx-lnx/(2sqrtx))/(8x)=#
#=(2-lnx)/(16xsqrtx)#
#f^((2))(x)=(-1/x*16xsqrtx-16*3/2*sqrtx*(2-lnx))/(16^2x^3)=#
#=(-16sqrtx-24sqrtx(2-lnx))/(16^2x^3)=#
#=(-8sqrtx(2+6-3lnx))/(16^2x^3)=(-sqrtx(8-3lnx))/(32x^3)#

The first derivative's zeros are stationary points:

#f'(x_s)=0 <=> (2-lnx_s)/(16x_ssqrtx_s)=0 <=>#
#(2-lnx_s=0 ^^ 16x_ssqrtx_s!=0) <=>#
#(lnx_s=2 ^^ x_s!=0) <=> (x_s=e^2 ^^ x_s!=0) <=> x_s=e^2#
#AAx>x_s: lnx>2 ^^ 2-lnx<0 ^^ 16xsqrtx>0#
#AAx>x_s : f'(x)<0#, #f# is decreasing
#AAx< x_s ^^ x>0 : lnx<2 ^^ 2-lnx>0 ^^ 16xsqrtx>0#
#AAx< x_s ^^ x>0 : f'(x)>0#, #f# is increasing
For #x=x_s# function #f# has maximum value:
#f_max=f(e^2)=lne^2/(8sqrt(e^2))=2/(8e)=1/(4e)#

If the second derivative changes sign at those points and the function is continuous, the points of inflection are zeros of the second derivative:

#f^((2))=0 <=> (-sqrtx_i(8-3lnx_i))/(32x_i^3)=0 <=>#
#(-sqrtx_i(8-3lnx_i)=0 ^^ 32x_i^3!=0) <=>#
#(8-3lnx_i=0 ^^ x_i!=0) <=> (lnx_i=8/3 ^^ x_i!=0) <=>#
#(x_i=e^(8/3) ^^ x_i!=0) <=> x_i=e^(8/3)=e^2root(3)(e^2)#
So, #x_i=e^2root(3)(e^2)# is potential inflection point.
#AAx>x_i: 8-3lnx<0, f^((2))>0# #AAx< x_i ^^ x>0: 8-3lnx>0, f^((2))<0#
Indeed, #x_i# is inflection point:
#f(x_i)=f(e^(8/3))=(lne^(8/3))/(8sqrt(e^(8/3)))=(8/3)/(8e^(4/3))=1/(3eroot(3)(e))#
Note: #f# is defined for #AAx>0# because of #lnx#, so I used that fact.
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Answer 2

To find increasing and decreasing intervals, inflection points, minimum, and maximum for ( f(x) = \frac{\ln(x)}{8\sqrt{x}} ):

  1. Find the first derivative: ( f'(x) = \frac{1}{8x\sqrt{x}} - \frac{\ln(x)}{16x^{\frac{3}{2}}} ).
  2. Set ( f'(x) = 0 ) and solve for ( x ) to find critical points.
  3. Determine the intervals where ( f'(x) > 0 ) for increasing intervals and where ( f'(x) < 0 ) for decreasing intervals.
  4. Find the second derivative: ( f''(x) ).
  5. Determine the sign of ( f''(x) ) in the intervals between critical points to find inflection points.
  6. Check the behavior of the function at the critical points and endpoints to find minimum and maximum points.
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Answer 3

To find increasing, decreasing, inflection points, minimum, and maximum for the graph of ( f(x) = \frac{\ln(x)}{8\sqrt{x}} ):

  1. Increasing and Decreasing Intervals:

    • ( f(x) ) is increasing when its derivative is positive.
    • ( f(x) ) is decreasing when its derivative is negative.
  2. First Derivative:

    • Find the first derivative of ( f(x) ) using the quotient rule.
    • ( f'(x) = \frac{d}{dx} \left( \frac{\ln(x)}{8\sqrt{x}} \right) )
  3. Second Derivative:

    • To find inflection points, we need the second derivative.
    • ( f''(x) = \frac{d^2}{dx^2} \left( \frac{\ln(x)}{8\sqrt{x}} \right) )
  4. Critical Points:

    • Critical points occur where ( f'(x) = 0 ) or ( f'(x) ) is undefined.
    • Solve ( f'(x) = 0 ) to find critical points.
  5. Minimum and Maximum Points:

    • Use the first derivative test or the second derivative test to determine minimum and maximum points.
    • If ( f''(x) > 0 ), it's a minimum point.
    • If ( f''(x) < 0 ), it's a maximum point.
  6. Procedure Summary:

    • Find ( f'(x) ) and ( f''(x) ).
    • Determine where ( f'(x) = 0 ) and any points where ( f'(x) ) is undefined.
    • Test intervals between critical points using the first or second derivative test.
    • Identify increasing/decreasing intervals and minimum/maximum points.

Following this procedure will help in analyzing the function ( f(x) = \frac{\ln(x)}{8\sqrt{x}} ) for its increasing/decreasing behavior, inflection points, and minimum/maximum points.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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