How do you find an equation of the tangent line to the graph of #f(x) = e^(x/2) ln(x) # at its inflection point?

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Answer 1 · 2016-05-05T13:59:19

Find the inflection point, then find the equation of the tangent at that point.

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

Use the product rule to find #f'(x)#.

#f'(x) = 1/2e^(x/2) ln(x) +e^(x/2) 1/x#

# = (xe^(x/2)lnx+2e^(x/2))/(2x)#

# = (e^(x/2)(xlnx+2))/(2x)#

Now use the quotient and product rules to find #f''(x)#.

(Details omitted)

#f''(x) = (e^(x/2)(x^2lnx+4x-4))/(4x^2)#

Find point(s) of inflection
The denominator and the factor #e^(x/2)# are always positive, so the sign of #f''# depends only on the sign of

#x^2lnx+4x-4#

We do not have an algebraic algorithm for solving

#x^2lnx+4x-4 = 0#

But observe that at #x=1#, we have #4x-4 = 0#.

Furthermore, at #x=1# we know #lnx = 0#.

Therefore #x=1# is a solution.

Note that the domain of #f# is #(0,oo)#

If #0 < x < 1#, then #x^2lnx# is negative and #4x-4# is also negative. The sum of two negatives is negative. Therefore, #f''(x) < 0# for #x < 1#

If #x > 1#, then #x^2lnx# is positive and #4x-4# is also positive. The sum of two positives is positive. Therefore, #f''(x) > 0# for #x > 1#

This assures us that there cannot be two inflection points, and, since #f(1) = 0#, we find that #(1,0)# is the only point of inflection.

Find the equation of the tangent line

Recall (from above) that #f'(x) = (e^(x/2)(xlnx+2))/(2x)#

At #(1,0)#, the slope of the tangent is #f'(1) = e^(1/2) = sqrte#

The equation of the line through #(1,0)# with slope #m=sqrte# is

#y=sqrte(x-1)#

(Put this in general, standard or slope intercept form if needed.)