# Is #f(x)=cosx# concave or convex at #x=(3pi)/2#?

See below.

We can determine where a function is convex or concave, by using the second derivative. If:

The second derivative is just the derivative of the first derivative. .i.e.

For

Now we solve the inequalities:

Notice that

This is verified by its graph:

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To determine if ( f(x) = \cos(x) ) is concave or convex at ( x = \frac{3\pi}{2} ), we need to check the second derivative of the function at that point.

First, find the second derivative of ( f(x) = \cos(x) ):

[ f'(x) = -\sin(x) ] [ f''(x) = -\cos(x) ]

Now, evaluate the second derivative at ( x = \frac{3\pi}{2} ):

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

Since the cosine function is negative at ( \frac{3\pi}{2} ), ( f''\left(\frac{3\pi}{2}\right) ) is negative.

A function is concave if its second derivative is negative, and convex if its second derivative is positive. Therefore, at ( x = \frac{3\pi}{2} ), the function ( f(x) = \cos(x) ) is concave.

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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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