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Is #f(x)=(x-3)(x+2)(x-4)^2# concave or convex at #x=-1#?

Answer 1

Isaiah Allen

The function is convex at #x=-1#

We need

#(uvw)'=u'vw+v'uw+w'uv#

We must calculate the first and second derivatives

#f(x)=(x-3)(x+2)(x-4)^2#

#f'(x)=(x+2)(x-4)^2+(x-3)(x-4)^2+2(x-3)(x+2)(x-4)#

#=(x-4){(x+2)(x-4)+(x-3)(x-4)+2(x-3)(x+2)}#

#=(x-4){x^2-2x-8+x^2-7x+12+2x^2-2x-12}#

#=(x-4)(4x^2-11x-8)#

#f''(x)=(4x^2-11x-8)+(x-4)(8x-11)#

#=4x^2-11x-8+8x^2-43x+44#

#=12x^2-54x+36#

Therefore,

#f''(-1)=12+54+36=102#

As, #f''(-1)>0#, we conclude that the function is convex

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Answer 2

Evelyn Agard

To determine the concavity of ( f(x) = (x-3)(x+2)(x-4)^2 ) at ( x = -1 ), we need to evaluate the second derivative of ( f(x) ) at ( x = -1 ). If the second derivative is positive, the function is concave up at that point. If it is negative, the function is concave down. If the second derivative is zero, the concavity is undefined.

First, find the first derivative of ( f(x) ): [ f'(x) = (x+2)(x-4)^2 + (x-3)(2(x-4)(x+2)) ]

Now, find the second derivative of ( f(x) ): [ f''(x) = (x-4)^2 + 2(x-4)(x+2) + (x+2)(2(x-4)) + (x-3)(2(x+2)) ]

Now, evaluate ( f''(-1) ) to determine the concavity 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.