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Is #f(x)=1-xe^(-3x)# concave or convex at #x=-2#?

Answer 1

Ezra Adams

function is convex at #x=-2#

Given function:

#f(x)=1-xe^{-3x}#

#f'(x)=-x(-3e^{-3x})-e^{-3x}(1)#

#f'(x)=e^{-3x}(3x-1)#

#f''(x)=e^{-3x}(3)+(3x-1)(-3e^{-3x})#

#f''(x)=e^{-3x}(6-9x)#

#f''(-2)=e^{6}(6-9(-2))#

#=24e^{6}#

Since #f(-2)>0# hence the given function is convex at #x=-2#

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Answer 2

Cameron Alexander

To determine whether ( f(x) = 1 - xe^{-3x} ) is concave or convex at ( x = -2 ), we need to find the second derivative of ( f(x) ) and then evaluate it at ( x = -2 ). If the second derivative is positive at ( x = -2 ), the function is concave upward (convex), and if it's negative, the function is concave downward.

The first derivative of ( f(x) ) is: [ f'(x) = e^{-3x}(3x - 1) ]

The second derivative of ( f(x) ) is: [ f''(x) = 9xe^{-3x} - 6e^{-3x} ]

Substituting ( x = -2 ) into ( f''(x) ): [ f''(-2) = 9(-2)e^{-3(-2)} - 6e^{-3(-2)} ]

Solving this expression: [ f''(-2) = -18e^{6} - 6e^{6} ]

[ f''(-2) = -24e^{6} ]

Since ( -24e^{6} ) is negative, the function ( f(x) ) is concave downward at ( x = -2 ).

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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.