# Determine the intervals on which the function f(x)=x^x defined on the interval (0,∞) is decreasing and increasing.

lemjay | Certified Educator

`y = x^x`

Note that we will not be able to apply both the power rule and exponential rule for derivatives since the above expression is neither in the form x^n and b^x.

To be able to apply the rules for derivative, take the logarithm of both sides.

`ln y = ln x^x`

`ln y = x ln x`

Then, perform implicit differentiation.

`1/y (dy/dx) = x*1/x + lnx * 1`

`1/y y' = 1 + ln x`

`y' = y(1 + ln x)`

Express right side as one variable. Note y = x^x.

`y' = x^x(1+ln x)`

Then, set y' to zero.

`0 = x^x (1+lnx)`

Set each factor to zero.

`1 + ln x = 0`

`ln x = -1`

`x = e^-1 = 1/e`

`x^x = 0`

`ln x^x = ln 0 `           ==> Note that in logarithm ( `log_b a = m` or `ln a = m` ), `a` must be greater than zero. Hence, we cannot consider this factor.

So we only have one critical point, `x=1/e` .

This means that we have two intervals which are less than 1/e and greater than 1/e. To determine at what interval is the function is increasing or decreasing, assign values of x for each interval. Then, substitute it to y'. Note that if the value of y' is positive, the function is increasing at that interval. If y' is negative, then the function is decreasing at that interval.

If `xlt1/e` , set `x =1/4`

`y' = (1/4)^(1/4)[1+ln(1/4)] = (0.707)(-0.386) =-0.273 `

y' is negative, thus the function is decreasing.

If` xgt1/e` , set `x=1`

` y' = 1^1 ( 1 + ln1) =1 (1) = 1`

y' is positive, hence the function is increasing.

Since the change of sign before and after x=1/e, is from negative to positive, then we have a minimum point. Note that, we do not have a maximum point since there is only one value of x, when y' is set to zero.

To determine the minimum point, substitute x=1/e to y.

`y = x^x = (1/e)^(1/e)= 1/e^(1/e)=0.69`

The minimum point is (1/e, 0.69)

Therefore at the interval (0,`oo` ), the function is decreasing at interval (0,`1/e` ). The function is increasing at (`1/e` ,`oo` ). Also at interval (0,`oo` ), the function has a local minima at point(`1/e`,0.69) and no local maxima.

justaguide | Certified Educator

The function `f(x) = x^x` . The interval in which f(x) is increasing is one in which f'(x) > 0 and the function is decreasing in the interval where f'(x) is negative.

`f'(x) = x^x*ln(x) + x^x = x^x*(ln x + 1)`

ln x is defined only for values of x > 0. In this case x^x is always positive.

The sign of f'(x) depends only on ln x + 1

ln x + 1 > 0

=> ln x > -1

=> x > 1/e

ln x + 1 < 0

=> ln x < -1

=> x < 1/e

The function `f(x) = x^x` is increasing in the interval `(1/e, oo)` and decreasing in the interval `(0, 1/e)`