On what interval is f concave upward? f(x)=xln(x)

#f(x)=xln(x)# is concave up when #f''(x)>0#

1 Answer
Oct 25, 2016

#f(x)=xln(x)# is concave up on the interval #(0,∞)#

Explanation:

To start off, we must realize that a function #f(x)# is concave upward when #f''(x)# is positive.

To find #f'(x)#, the Product Rule must be used and the derivative of the natural logarithmic function must be known:

Product Rule (Simplified format of the rule): (Derivative of first)(Second function) + (First function)(Derivative of the second function)
Derivative of the natural logarithmic function: #d/dx (lnx)= 1/x#

#f'(x)= (1*lnx)+(x*1/x)=lnx+1#

Now we need to take the derivative of this in order to find #f''(x)#. Remember that the derivative of a constant is equal to zero.

#f''(x)=1/x#

Remember that this question is asking us to look for when the function is concave up, meaning where #f''(x)>0#.

#f''(x) = 1/x > 0#
Since #1# is a positive value, #1/x# is positive whenever #x>0# (a fraction is positive if the numerator and denominator have the same sign).

The final answer is that the function #f(x)=xlnx# is concave up on the interval #(0,∞)#, which is when #x>0#.