Determine the simplified form for the slope of the tangent at any point for the function p(x)=1/x+1
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.
Sorry I meant 1/(x + 1), thank you very much for your help!!!
To determine the slope of the tangent at any point for the function p(x) = 1/(x + 1), we need to find the derivative of the function and then simplify it.
Step 1: Find the derivative
To find the derivative, we can use the power rule for differentiation. Let's apply it to the function p(x).
p(x) = 1/(x + 1)
Using the power rule, we differentiate by bringing down the power in front and reducing the power by 1.
p'(x) = -1/(x + 1)^2
Step 2: Simplify the derivative
To simplify the derivative, we can start by getting rid of the negative sign by multiplying both the numerator and denominator by -1.
p'(x) = -1/(x + 1)^2
Multiplying the numerator and denominator by -1 gives:
p'(x) = 1/-(x + 1)^2
Simplifying further, we can move the negative sign to the numerator.
p'(x) = -1/(x + 1)^2
So, the simplified form for the slope of the tangent at any point for the function p(x) = 1/(x + 1) is -1/(x + 1)^2.