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Is #f(x)=e^(2x)-x^2/(x-1)-1# concave or convex at #x=0#?

Answer 1

f(x) is convex at x=0

Let's calculate the first and the second derivative of f(x):

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

#f''(x)=4e^(2x)-((2x-2)(x-1)^2-2(x-1)(x^2-2x))/(x-1)^4# #=4e^(2x)-(2(x-1)^(cancel3^2)-2cancel((x-1))(x^2-2x))/(x-1)^(cancel4^3)# #=4e^(2x)-(2(x-1)^2-2(x^2-2x))/(x-1)^3#

Then let's look at the sign of f''(0):

#f''(0)=4e^0-2/-1=4+2=6>0#

Then f(x) is convex at x=0

graph{e^(2x)-x^2/(x-1)-1 [-5.164, 5.17, -2.583, 2.58]}

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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.