Is #f(x)=1/(x-1)-1/(x+1)^2# increasing or decreasing at #x=0#?
Since
now use the chain rule:
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To determine if ( f(x) = \frac{1}{x-1} - \frac{1}{(x+1)^2} ) is increasing or decreasing at ( x = 0 ), we need to evaluate the sign of its derivative at that point.
The derivative of ( f(x) ) is ( f'(x) = \frac{-1}{(x-1)^2} + \frac{2}{(x+1)^3} ).
At ( x = 0 ), ( f'(0) = \frac{-1}{(-1)^2} + \frac{2}{(1)^3} = -1 + 2 = 1 ).
Since the derivative is positive at ( x = 0 ), the function ( f(x) ) is increasing at that point.
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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.
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 extrema of #f(x)=-sinx-cosx# on the interval #[0,2pi]#?

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