For the function f(x) =1 over x squared, determine a general expression for the slope of a tangent using the limit as h approaches 0 of (Instantaneous rate of change equation) .Then, let a = 1. What does this mean (in words) when we let a = 1?
f(x) = 1/x^2
The slope of the tangent is the limit of
(f(x+h)-f(x))/h
f(x+h) = 1/(x+h)^2
so,
f(x+h)-f(x) = 1/(x+h)^2 - 1/x^2
= (x^2-(x+h)^2)/(x^2(x+h)^2)
= (x^2-x^2-2xh-h^2)/(x^2(x+h)^2)
= (-2xh-h^2)/(x^2(x+h)^2)
Now divide that by h and you have
(-2x-h)/(x^2(x+h)^2)
Now take the limit as h->0 and you have
-2x/x^4 = -2/x^3
Now evaluate that at x=a and the slope at (1,1) = -2
They used a because (if you check your text)
f'(a) = limit (f(a+h)-f(a))/h