#### To determine

**To find:**

Intervals of concavity and the inflection points of the function f(x)=(lnx)x

#### Answer

Concave up in the interval (e83,∞), concave down in the interval (0,e83) and the point of inflection is (e83,83e−43)

#### Explanation

**Given:**

The function, f(x)=(lnx)x

**Formulae used:** ddx(f(x)g(x))=g(x)f′(x)−f(x)g′(x)[g(x)]2

The function is f(x)=(lnx)x

Use the quotient rule of differentiation ddx(f(x)g(x))=g(x)f′(x)−f(x)g′(x)[g(x)]2 to evaluate the given function as follows:

f(x)=lnxxf′(x)=x1x−lnx12xx=2−lnx2x32

Differentiate again with respect to *x* and get:

f″(x)=2x32(−1x)−(2−lnx)(3x12)4x3=3lnx−84x52>0lnx>83x>e83

Take a point x=2 in the interval (0,e83). The value of f″(x) at this point is calculated as follows:

3lnx−84x52=3ln(2)−84(2)52=2.079−822.627=−0.262<0

Since, f″(x)<0 in the interval (0,e83).

Therefore, in the interval (0,e83), the function f(x) is concave down.

Now, take a point x=15 in the interval (e83,∞). The value of f″(x) at this point is calculated as follows:

3lnx−84x52=3ln(15)−84(15)52=0.1243485.685=3.557×10−5>0

Since, f″(x)>0 in the interval (e83,∞).

Therefore, in the interval (e83,∞), the function f(x) is concave up.

Thus, it can be observed that f is concave upward on (e83,∞), concave downward on (0,e83).

Therefore, the inflection point is (e83,83e−43).

**Conclusion:**

Hence, the inflection point of the function f(x)=(lnx)x is (e83,83e−43).