# How do you determine whether the function #f(x)= (lnx)^2# is concave up or concave down and its intervals?

##### 1 Answer

#### Answer:

Concave up on

#### Explanation:

The concavity of a function is determined by the sign of the second derivative of the function:

- If
#f''(a)<0# , then#f(x)# is concave down at#x=a# . - If
#f''(a)>0# , then#f(x)# is concave up at#x=a# .

Find the second derivative of the function. But first, we must find the first derivative, which will require the chain rule:

#f'(x)=d/dx[lnx]*2lnx=1/x(2lnx)=(2lnx)/x#

To differentiate this, use the quotient rule.

#f''(x)=(xd/dx[2lnx]-2lnxd/dx[x])/x^2=(x(2/x)-2(1)lnx)/x^2=(2-2lnx)/x^2#

Now, to determine the intervals of concavity, we have to find when the second derivative is positive and when it's negative.

The sign of the second derivative could change, that is, go from positive or negative or vice versa, when it's equal to *possible* point where the concavity could shift (called a possible point of inflection).

#(2-2lnx)/x^2=0#

#2-2lnx=0#

#lnx=1#

#x=e#

Thus the concavity could shift when

**When #mathbf(0< x < e)#:**

#f''(1)=(2-2ln(1))/1^2=(2-0)/1=2#

Since this is positive, we know that

**When #mathbf(x>e)#:**

#f''(e^2)=(2-2ln(e^2))/(e^2)^2=(2-2(2))/e^4=-2/e^4#

This is obviously negative. Since it is negative, we know this whole interval will be negative. Thus,

We can check a graph of

graph{((lnx)^2-y)((x-e)^2+(y-1)^2-.01)=0 [-4.52, 15, -1.314, 6.586]}