Is #f(x)=xlnx-x# concave or convex at #x=1#?

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Answer 1 · 2016-02-17T00:11:57

#f(x)# is convex at #x=1#.

#f(x)# is concave upward (convex) at a point #x_0# if #f''(x_0) > 0# and concave downward (concave) at a point #x_0# if #f''(x_0) < 0#.

In this case, we have

#f''(x) = d/dxf'(x)#

#=d/dx(d/dxxln(x)-x)#

#=d/dxln(x)#

#=1/x#

Then, at #x=1#:

#f'''(1) = 1/1 = 1 > 0#

Thus #f(x)# is convex at #x=1#.