How do you find the inflection point and concavity of #f(x)=x^(4)*ln (x)#?

Question

Christopher Agresta · Accepted Answer

#f_min = f( e^(-1/4)) approx -0.092#.
#f(x)# is concave up.

#f(x) = x^4ln(x)# Since #lnx# is defined for #x>0 -> f(x)# is defined for #x >0# To find an inflection point we need to find where #f'(x)=0# #f'(x) = 4x^3ln(x) + x^4*1/x# [Chain rule] #= 4x^3ln(x) + x^3# #f'(x)=0 -> 4x^3ln(x) + x^3 =0# #=>x^3(4lnx +1) =0# #:. x=0 or (4lnx +1)=0# Since #x>0# consider #4lnx = -1# #=>lnx = -1/4# #x= e^(-1/4)# #x approx 0.7788# Now, let's look at the graph of #f(x)# below. graph{ x^4ln(x) [-0.234, 1.452, -0.2815, 0.561]} We can see that #f(e^(-1/4))# is a minimum value of #f(x) approx -0.092#. #f(x)# is concave up at this point. From the nature of #f(x)# we may conclude that this is the only inflection point.

Emma Aldridge · Answer

undefined To find the inflection point and concavity of $ f(x) = x^4 \cdot \ln(x) $, follow these steps:

1. Find the first derivative of $ f(x) $ using the product rule: 
$$ f'(x) = 4x^3 \cdot \ln(x) + x^3 \cdot \frac{1}{x} $$

2. Simplify the derivative:
$$ f'(x) = 4x^3 \cdot \ln(x) + x^2 $$

3. Find the second derivative of $ f(x) $:
$$ f''(x) = 12x^2 \cdot \ln(x) + 8x^2 + 2x $$

4. Determine the points where the second derivative equals zero or is undefined to find potential inflection points.

5. Test the concavity by evaluating $ f''(x) $ at intervals determined by the critical points found in step 4.

6. Analyze the sign of $ f''(x) $ in each interval to determine the concavity of the function.

7. The inflection points occur at the values of $ x $ where the concavity changes.

8. The concavity of the function $ f(x) $ can be described as concave up or down based on the sign of $ f''(x) $ in the intervals.