Is the function concave up or down if #f(x)= (lnx)^2#?

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Answer 1 · 2015-04-13T17:55:56

It is concave up on the interval #(0,e)# and concave down on #(e, oo)#.

#f(x)=(lnx)^2#.

#f'(x) = 2(lnx)*1/x = (2lnx)/x#.

#f''(x) = ((2/x)*x - 2lnx *1)/x^2 = (2(1-lnx))/x^2#.

#1-lnx = 0# where #lnx = 1# which is at #x=e#.

All other factors of #f''(x)# are always positive, so the sign of #f''(x)# is the same as the sign of #1-lnx#.

That is:
#f''(x)# is positive if #x < e# (so #lnx < 1#) and
it is negative for #x > e# (where #lnx > 1 #).