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

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Answer 1 · 2015-03-10T18:21:33

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)#