# How do you find the linearization at x=0 of # f ' (x) = cos (x^2)#?

Use the fact that the linearization at

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graph{y=cos(x^2) [-4.15, 4.62, -1.61, 2.775]}

graph{(y-cos(x^2))(y-1)=0 [-4.15, 4.62, -1.61, 2.775]}

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To find the linearization of ( f'(x) = \cos(x^2) ) at ( x = 0 ), we use the formula for linearization:

[ L(x) = f'(a) + f''(a)(x - a) ]

Where ( a = 0 ). First, we find the first and second derivatives of ( f'(x) ).

[ f''(x) = -2x\sin(x^2) ]

Now, we evaluate these derivatives at ( x = 0 ):

[ f'(0) = \cos(0) = 1 ] [ f''(0) = -2(0)\sin(0) = 0 ]

Plug these values into the linearization formula:

[ L(x) = 1 + 0(x - 0) ] [ L(x) = 1 ]

So, the linearization of ( f'(x) = \cos(x^2) ) at ( x = 0 ) is ( L(x) = 1 ).

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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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