Answer:
The value of [tex]f(1.2)=y-f'(1.2)(x-1.2)[/tex].
Step-by-step explanation:
According to the given problem the value of taken functions are,
[tex]f(1)=3[/tex] and [tex]f'(1)=-2[/tex]
Then to estimate the value of f(t) at t=1.2, one can consider the equation of tangent line of the graph of f(x) at a point x=a is,
[tex]y=f(a)+f'(a)(x-a)\hfill (1)[/tex]
wnere f(x) is differentiable at the point x=a.
From (1) we can get,
[tex]f(a)=y-f'(a)(x-a)[/tex]
In the present problem a=1, and we know value of f(1) and f'(1).
Hence, at a=1.2 and after finding value of f'(1.2) one can easily find,
[tex]f(1.2)=y-f'(1.2)(x-1.2)[/tex]
which will give the best approximation.
