# For what values of x is #f(x)= -x^2-x+4# concave or convex?

x = -1/2 with downward concavity

Concavity is expressed by the second derivative. To start, we must take the first derivate to find the POINT(S):

What we just found are the extrema which we can simplify to mean: set y = 0 0 = -2x - 1 2x = -1 x = -1/2

To solidify this answer, take a look at the graph. graph{-x^2 - x + 4 [-10, 10, -5, 5]}

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To determine where the function (f(x) = -x^2 - x + 4) is concave or convex, we need to analyze its second derivative. If (f''(x) > 0), the function is convex, and if (f''(x) < 0), the function is concave.

Given (f(x) = -x^2 - x + 4), find the second derivative: [f'(x) = -2x - 1] [f''(x) = -2]

Since the second derivative (f''(x)) is a constant (-2), it's always negative. Thus, the function is concave for all real 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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- What are the points of inflection of #f(x)= 10(x-5)^3+2#?

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