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Determine the correct f(x) & a and evaluate L(x)=f(a)+f'(a)(x-a) for the value (sin(.01))^2.?
Please show the work used to come to the answer.
1 Answer
- 4 years ago
I'm going to do one more and then you're going to have to try some of these on your own.
f(x) = sin(x)^2
f'(x) = 2 * sin(x) * cos(x) = sin(2x)
sin(0.01)^2 is going to be pretty close to sin(0)^2
f(0) = sin(0)^2 = 0^2 = 0
f'(0) = sin(2 * 0) = 0
We need a line that passes through (0 , 0) with a slope of 0. y = 0
sin(0.01)^2 is going to be really close to 0.
9.9996666711110793652204581262368... * 10^(-5)
10 * 10^(-5)
10^(-6)
So we're right, within 1 part in a million.
