What is the local linearization of e^sin(x)esin(x) near x=1?

1 Answer
Aug 4, 2017

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(x) about x=1x=1 is given by the linear terms of the Taylor Series about the point x=ax=a, ie the terms given by:

f(x) ~~ f(a) + f'(a)(x-a)

Differentiating f(x) using the chain rule, we have:

f'(x) = e^(sinx) (cosx)

So with x=1 we have:

f(1) = e^(sin1)
" " = 2.3197768 ...

f'(1) = e^(sin1) (cos1)
" " = 1.2533897 ...

Hence, the linear approximation near x=1 is given by:

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 x=1.01 which is near x=1, Then:

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 3dp