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

Answer 1

#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#.

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Answer 2

Aurora Alvarado

To determine whether ( f(x) = x \ln x - x ) is concave or convex at ( x = 1 ), we need to find the second derivative of ( f(x) ) and then evaluate it at ( x = 1 ).

The first derivative of ( f(x) ) is: [ f'(x) = \frac{d}{dx}(x \ln x - x) = \ln x + 1 - 1 = \ln x ]

The second derivative of ( f(x) ) is: [ f''(x) = \frac{d}{dx}(\ln x) = \frac{1}{x} ]

Evaluating ( f''(1) ): [ f''(1) = \frac{1}{1} = 1 ]

Since the second derivative ( f''(1) = 1 ) is positive, ( f(x) ) is convex at ( x = 1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.