Which of the following is the best linear approximation for f(x) = sin(x) near x = π seconds?
sin π = 0
slope = cos π = -1
f(x) = sin(π) -1(x-π)
= -(x-π)
To find the best linear approximation for \(f(x) = \sin(x)\) near \(x = \pi\) seconds, we can use the concept of linearization. The linearization of a function at a point is given by the equation:
\(L(x) = f(a) + f'(a)(x-a)\)
where \(a\) is the point at which we want to approximate, \(f'(a)\) is the derivative of the function evaluated at \(a\), and \(L(x)\) is the linear approximation.
In this case, we want to find the linear approximation of \(f(x) = \sin(x)\) near \(x = \pi\) seconds. Let's start by finding the first derivative of \(f(x)\):
\(f'(x) = \cos(x)\)
Now, evaluate the derivative at \(x = \pi\):
\(f'(\pi) = \cos(\pi) = -1\)
Now we can construct the linear approximation by plugging in the given values:
\(L(x) = f(\pi) + f'(\pi)(x-\pi)\)
\(L(x) = \sin(\pi) + (-1)(x-\pi)\)
Since \(\sin(\pi) = 0\), the linear approximation becomes:
\(L(x) = -x + \pi\)
Therefore, the best linear approximation for \(f(x) = \sin(x)\) near \(x = \pi\) seconds is \(-x + \pi\).