Asked by ger
Determine the simplified form for the slope of the tangent at any point for the function p(x)=1/x+1
Answers
Answered by
Reiny
For some weird reason I will assume you meant
p(x) = 1/(x + 1)
p(x+h) = 1/(x+h+1)
slope of tangent at any point
= lim ( p(x+h) - p(x) )/h , as h ----> 0
= lim (1/(x+h+1) - 1/(x+1) )/h
= lim ( x+1 - x - h - 1)/((x+1)(x+h+1) ) / h
= lim ( -h/((x+1)(x+h+1) ) / h
= lim -1/((x+1)(x+h+1) , as h ---> 0
= -1/(x+1)^2
If you meant it the way you typed it, then just make the necessary changes.
p(x) = 1/(x + 1)
p(x+h) = 1/(x+h+1)
slope of tangent at any point
= lim ( p(x+h) - p(x) )/h , as h ----> 0
= lim (1/(x+h+1) - 1/(x+1) )/h
= lim ( x+1 - x - h - 1)/((x+1)(x+h+1) ) / h
= lim ( -h/((x+1)(x+h+1) ) / h
= lim -1/((x+1)(x+h+1) , as h ---> 0
= -1/(x+1)^2
If you meant it the way you typed it, then just make the necessary changes.
Answered by
ger
Sorry I meant 1/(x + 1), thank you very much for your help!!!
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.