Asked by Nate
The linear approximation to (1/sqrt(9-x)) at x=0. What is y?
Answers
Answered by
Steve
the linear approximation is just the tangent line at the given point. Within a suitably small interval, the line approximates the curve to any desired accuracy.
So,
y = 1/√(9-x)
y' = 1 / 2(9-x)^(3/2)
at (0,1/3), y' = 1 / 6√3
So, now we have a point and a slope. The line is
y - 1/3 = (1/6√3)(x-0)
So,
y = 1/√(9-x)
y' = 1 / 2(9-x)^(3/2)
at (0,1/3), y' = 1 / 6√3
So, now we have a point and a slope. The line is
y - 1/3 = (1/6√3)(x-0)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.