How do you use linear approximation to estimate #g(2.95)# and #g(3.05)# if you know that #g(3)=-5#?

Answer 1
With the given information, the best you can do is: #g(2.95)~~-5 - 0.05g'(3)# and #g(3.05)~~-5 + 0.05g'(3)#
The linear approximation for #g(x)# near #3# is the equation of the line tangent to the graph of #g# at the point #(3, g(3))=(3, -5)#.
It is: #g(x)~~g(3)+g'(3)(x-3)#.
Because you know #g(3)#, but not #g'(3)#, the best you can do is:
#g(x)~~-5+g'(3)(x-3)#.
For #x=2.95#, we get #x-3=-0.05# And for #x=3.05#, we get #x-3=0.05#.
So, #g(2.95)~~-5 - 0.05g'(3)# and #g(3.05)~~-5 + 0.05g'(3)#
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Answer 2

To estimate ( g(2.95) ) and ( g(3.05) ), use linear approximation:

  1. Find the slope of the function at ( x = 3 ) using the derivative ( g'(x) ).
  2. Use the point-slope form of a linear equation to find the linear approximation.
  3. Plug in ( x = 2.95 ) and ( x = 3.05 ) into the linear equation to estimate ( g(2.95) ) and ( g(3.05) ).
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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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