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

Answer 1

Luke Alvarez

It is convex.

If #f''(-1)<0#, the function is concave at that point. If #f''(-1)>0#, the function is convex at that point.

#f'(x)=-9x^2+8x+3#

#f''(x)=-18x+8#

#f''(-1)=26#

Since #26>0#, the function is convex at the point #(-1,0)#.

graph{-3x^3+4x^2+3x-4 [-9.96, 10.04, -6.12, 3.88]}

The graph does appear to be convex (forming a #uu# shape) at the specified point.

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Answer 2

Isabella Ackerman

To determine the concavity or convexity of ( f(x) = -3x^3 + 4x^2 + 3x - 4 ) at ( x = -1 ), we need to examine the sign of the second derivative, ( f''(x) ), at that point.

First, find the first derivative ( f'(x) ) by differentiating ( f(x) ): [ f'(x) = -9x^2 + 8x + 3 ]

Now, find the second derivative ( f''(x) ) by differentiating ( f'(x) ): [ f''(x) = -18x + 8 ]

Evaluate ( f''(-1) ): [ f''(-1) = -18(-1) + 8 = -18 + 8 = -10 ]

Since ( f''(-1) = -10 ) is negative, the function is concave down 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.