Question

HELP!!! Which of the following is the best linear approximation for f(x) = sin(x) near x = π seconds?

y = –x + π − 1
y = –1
y = –x + π
y = –x − π

Answer 1

answer should be

y = –x + π

f(π) + f'(π)(x - π)
but
f(π) = 0 and f'(π) = -1
so
f(π) + f'(π)(x - π)
= 0 + -1(x - π)
= -x + π

hope it helps

Answer 2

y = –x + π

Explanation:

We have been given that

f(x) = sin (x)

On differentiating, we get

f'(x) = cos (x)

Linear approximation near any point 'a' is given by the formula

`L(x)\approx f(a)+f'(a)(x-a)`

here, a = π

Hence, f(π) = sin (π) = 0

And f'(π) = cos(π) =-1

Substituting these values in the above formula, we get

`L(x)\approx 0+(-1)(x-\pi)\\\\L(x)\approx 0-x+\pi\\\\L(x)\approx -x+\pi`

Therefore, the linear approximation for f(x) = sin(x) near x = π seconds is y = –x + π

Third option is correct.