What is the local linearization of e^sin(x)esin(x) near x=1?
1 Answer
f(x) ~~ 1.2534x+1.0664 f(x)≈1.2534x+1.0664
Explanation:
Let:
f(x) = e^(sinx) f(x)=esinx
The linear approximation of a function
f(x) ~~ f(a) + f'(a)(x-a)
Differentiating
f'(x) = e^(sinx) (cosx)
So with
f(1) = e^(sin1)
" " = 2.3197768 ...
f'(1) = e^(sin1) (cos1)
" " = 1.2533897 ...
Hence, the linear approximation near
f(x) ~~ 2.3198 + 1.2534(x-1)
" " = 2.3198 + 1.2534x-1.2534
" " = 1.2534x+1.0664
Where we have rounded to 4dp.
Example:
Consider the case
f(1.01) = e^(sin(1.01))
" " = 2.332246 ...
And the linear approximation gives us:
f(1.01) ~~ 1.2534(1.01)+1.0664
" " = 1.265934+1.0664
" " = 2.332334
Which is correct within