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?

Question

Scarlett Allison · Accepted Answer

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.)

Axel Alvarez · Answer

undefined To find the equation of the tangent line to the graph of f(x) = e^(x/2) ln(x) at its inflection point, we need to follow these steps:

1. Find the second derivative of f(x) by differentiating twice.
2. Set the second derivative equal to zero and solve for x to find the x-coordinate of the inflection point.
3. Substitute the x-coordinate of the inflection point into the original function f(x) to find the corresponding y-coordinate.
4. Use the point-slope form of a line to find the equation of the tangent line, using the slope of the tangent line as the derivative of f(x) evaluated at the inflection point.

Let's go through these steps:

1. Differentiating f(x) once, we get f'(x) = (1/2)e^(x/2) ln(x) + e^(x/2)/x.
2. Differentiating f'(x) again, we get f''(x) = (1/4)e^(x/2) ln(x) + (1/4)e^(x/2)/x + e^(x/2)/x^2.
3. Setting f''(x) equal to zero, we solve the equation (1/4)e^(x/2) ln(x) + (1/4)e^(x/2)/x + e^(x/2)/x^2 = 0 to find the x-coordinate of the inflection point.
4. Substitute the x-coordinate of the inflection point into f(x) to find the corresponding y-coordinate.
5. Calculate the derivative of f(x) and evaluate it at the inflection point to find the slope of the tangent line.
6. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the inflection point and m is the slope, to find the equation of the tangent line.