Question

f(x)=1/ (3-2x) ; P (0,1)
A) use definition to find the derivitive to the graph of f at P.
b) Determine an Equation of the tangent line at P

Answer 1

A) the definition of lim h-> 0 f'(x) = [f(x+h) - f(x)] / h so it's a matter of plug and chug.
f(x+h) = 1 / (3 - 2 (x+h)) = 1 / (3 - 2x - 2h)
f(x) = 1/ (3-2x)
f(x+h) - f(x) = 1 / (3 - 2x - 2h) - 1/ (3-2x)
Multiply 1 / (3 - 2x - 2h) by (3-2x)/(3-2x) and 1/ (3-2x) by (3 - 2x - 2h)/(3 - 2x - 2h) to get common denominators
you now have [(3-2x) - (3 - 2x - 2h)]/[(3 - 2x - 2h)(3-2x)]
Simplify to get -2h/(9 -6x - 6h -6x + 4x^2 +4hx)
simplify further and you get -2h/(9-12x -6h +4hx +4x^2)
Now you add in the last part, divide by h.
Your equation is now (-2h/(9-12x -6h +4hx +4x^2))/h =
-2/(9-12x -6h +4hx +4x^2)
Now insert the limit as h approaches 0 and you get:
-2/(9-12x +4x^2)

Answer 2

B) the equation of the tanget is essentially y=mx + b where m = f'(0) and your b = f(0)
f'(0) = -2/(9-12(0) +4(0)^2) = -2/9
b = f(0) = 1/ (3-2(0)) = 1/3
y = (-2/9)x + 1/3 is the equation of the line tangent to f(x) = 1/ (3-2x) at point P(0,1)

Answer 3

ok