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

Question

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

Scarlett Allison · Accepted Answer

Concave up on #(0,e)#; concave down on #(e,+oo)#

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

Find the second derivative of the function. But first, we must find the first derivative, which will require the chain rule: #d/dx[u^2]=u'*2u#, and here #u=lnx#. Thus,

#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 #0#. Set #f''(x)=0# to find a 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 #x=e#. We can test the sign of the second derivative around this point.

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

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

Since this is positive, we know that #f(x)# is concave up on #(0,e)#. Note that the interval is not #(-oo,e)# since #lnx# is only defined for #x>0#.

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, #f(x)# is concave down on #(e,+oo)#.

We can check a graph of #f(x)#: the concave up portion should resemble the #uu# shape, and the concave down portion should resemble the #nn# shape. The small circle is where the concavity shifts, which is the point of inflection: #(e,1)#.

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

Samantha Adkison · Answer

undefined To determine the concavity of the function f(x) = (ln x)^2 and its intervals, we need to find its second derivative and then analyze its sign. First, find the first derivative of f(x): f'(x) = 2(ln x)(1/x) = 2ln x / x. Next, find the second derivative of f(x): f''(x) = d/dx [2ln x / x] = (1/x)(2/x) - 2ln x / x^2 = (2 - 2ln x) / x^2. To determine concavity, we need to find where f''(x) > 0 (concave up) and where f''(x) < 0 (concave down). Setting f''(x) > 0: (2 - 2ln x) / x^2 > 0 2 - 2ln x > 0 2 > 2ln x 1 > ln x e^1 > x x < e. So, the function is concave up for x < e, and concave down for x > e.