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.