# For what values of x is #f(x)=(-2x^2)/(x-1)# concave or convex?

The function will be concave in

The function will be convex in

Given function:

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To determine the concavity or convexity of ( f(x) = \frac{-2x^2}{x-1} ), we first find its second derivative, ( f''(x) ). Then, we analyze the sign of ( f''(x) ) to determine concavity or convexity.

First, we find the first derivative: [ f'(x) = \frac{d}{dx} \left( \frac{-2x^2}{x-1} \right) = \frac{-2x^2 \cdot (x-1) - (-2x^2) \cdot 1}{(x-1)^2} = \frac{-2x^2 + 2x^2}{(x-1)^2} = \frac{0}{(x-1)^2} = 0 ]

Since the first derivative equals zero, we need to apply the quotient rule to find the second derivative: [ f''(x) = \frac{d}{dx} \left( \frac{-2x^2}{x-1} \right) = \frac{-2(x-1)(2x^2) - (-2x^2)(1)}{(x-1)^2} = \frac{-4x^2 + 4x^2}{(x-1)^2} = 0 ]

Since the second derivative equals zero, the concavity or convexity of ( f(x) ) is undetermined at this point. We need to examine the points of inflection, which occur when ( f''(x) = 0 ) or is undefined.

Setting ( f''(x) = 0 ), we have: [ -4x^2 + 4x^2 = 0 ] [ 0 = 0 ]

This equation is satisfied for all real values of ( x ). Therefore, there are no points of inflection for ( f(x) ).

Thus, the function ( f(x) = \frac{-2x^2}{x-1} ) is neither concave nor convex for any values of ( x ).

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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.

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.

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.

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