Is #f(x)=sinx-cosx# increasing or decreasing at #x=0#?

Answer 1

Samantha Adkison

#f(x)=sinx-cosx# is increasing at #x=0#

Whether a function #f(x)# is increasing or decreasing at #x=0#, is determined by the value of its derivative at #x=0#.

As #f(x)=sinx-cosx#

#f'(x)=cosx-(-sinx)=cosx+sinx#

and at #x=0#, #f'(0)=cos0+sin0=1+0=1#

As #f'(0)>0#, #f(x)# is increasing at #x=0# graph{sinx-cosx [-5, 5, -2.5, 2.5]}

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Answer 2

Emma Aldridge

To determine if ( f(x) = \sin(x) - \cos(x) ) is increasing or decreasing at ( x = 0 ), we need to analyze the sign of the derivative of ( f(x) ) at ( x = 0 ).

The derivative of ( f(x) ) is ( f'(x) = \cos(x) + \sin(x) ).

Evaluate ( f'(0) = \cos(0) + \sin(0) = 1 + 0 = 1 ).

Since ( f'(0) > 0 ), the function ( f(x) = \sin(x) - \cos(x) ) is increasing at ( x = 0 ).

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Answer from HIX Tutor

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.