# Is #f(x) =cscx-sinx# concave or convex at #x=pi/3#?

God bless....I hope the explanation is useful.

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A concave function is a function in which no line segment joining two points on its graph lies above the graph at any point.

A convex function, on the other hand, is a function in which no line segment joining two points on the graph lies below the graph at any point.

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To determine if ( f(x) = \csc(x) - \sin(x) ) is concave or convex at ( x = \frac{\pi}{3} ), we need to evaluate the second derivative of ( f(x) ) at that point. If the second derivative is positive, the function is concave upward (convex), and if it is negative, the function is concave downward.

The second derivative of ( f(x) ) with respect to ( x ) is given by:

[ f''(x) = -\csc(x)\cot(x) - \cos(x) ]

At ( x = \frac{\pi}{3} ), we have:

[ \csc\left(\frac{\pi}{3}\right) = 2, \quad \cot\left(\frac{\pi}{3}\right) = \sqrt{3}, \quad \text{and} \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} ]

Plugging these values into the second derivative, we get:

[ f''\left(\frac{\pi}{3}\right) = -2\sqrt{3} - \frac{1}{2} ]

Since ( -2\sqrt{3} - \frac{1}{2} ) is negative, ( f(x) = \csc(x) - \sin(x) ) is concave downward at ( x = \frac{\pi}{3} ).

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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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