Question

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

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

Answer 1

`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`.