How do you find the linearization at x=1 of #y = 2/x#?

Answer 1

#L(x)=4-2x#

#L(x)=f(a)+f'(a)(x-a)#
Here, #f(x)=2/x=2x^-1#, #a=1, f(a)=2.#
So, #f'(x)=-2/x^2, f'(1)=-2#, #(x-a)=(x-1)#

The linearization is then

#L(x)=2-2(x-1)#
#L(x)=2-2x+2#
#L(x)=4-2x#
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Answer 2

To find the linearization at ( x = 1 ) of ( y = \frac{2}{x} ), follow these steps:

  1. Calculate the derivative of ( y = \frac{2}{x} ) to find the slope of the tangent line at ( x = 1 ).
  2. Evaluate the derivative at ( x = 1 ) to find the slope of the tangent line.
  3. Use the point-slope form of the equation of a line to write the equation of the tangent line.
  4. This equation represents the linearization at ( x = 1 ) of ( y = \frac{2}{x} ).
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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.

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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