How do you find the inflection point of the function #f(x) = x^2ln(x)#?

Question

Christopher Allers · Accepted Answer

The inflection is the zeroes of the second derivative.

The inflection point is #x=e^(-3/2)# Thus, the first derivative is available. #f'(x)=2xlnx+x^2*1/x=2xlnx+x# as well as the second derivative #f''(x)=2lnx+2x*1/x+1=> f''(x)=2lnx+3# Therefore, f''(x)'s zeroes are #f''(x)=0=>2lnx+3=0=>lnx^2=-3=>x=e^(-3/2)#

Charles Arocha · Answer

undefined To find the inflection point of the function $ f(x) = x^2 \ln(x) $, follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for $ x $.
3. Check the concavity of the function around the critical point found in step 2.

Let's proceed:

1. $ f(x) = x^2 \ln(x) $
   First derivative: $ f'(x) = 2x \ln(x) + x $
   Second derivative: $ f''(x) = 2 \ln(x) + \frac{2}{x} + 1 $

2. Setting the second derivative equal to zero:
   $ 2 \ln(x) + \frac{2}{x} + 1 = 0 $
   $ 2 \ln(x) + \frac{2}{x} = -1 $
   $ 2 \ln(x) = -1 - \frac{2}{x} $
   $ \ln(x) = -\frac{1}{2} - \frac{1}{x} $
   $ e^{\ln(x)} = e^{-\frac{1}{2} - \frac{1}{x}} $
   $ x = e^{-\frac{1}{2}} \cdot e^{-\frac{1}{x}} $
   $ x = e^{-\frac{1}{2}} \cdot e^{-\frac{1}{x}} $

3. We can use numerical methods or approximations to find the value of $ x $. Once you have found the value of $ x $, plug it back into the second derivative to determine the concavity of the function around the inflection point. If the second derivative changes sign from positive to negative or vice versa, then the point is an inflection point.