f(x+h) = 1/(x+h)^2
f(x) = 1/x^2
f(x+h) - f(x) = 1/(x+h)^2 - 1/x^2
=x^2/[x^2(x+h)^2] - (x+h)^2 /[x^2(x+h)^2]
=[ x^2 - (x+h)^2 ]/ [x^2(x+h)^2]
= [ - 2 x h - h^2 ]/[x^2(x+h)^2]
divide by h
= [ - 2 x - h ]/[x^2(x+h)^2]
let h ---> 0
=-2x/x^4
= -2/x^3
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 IROC equation
2 answers
f(x)=1/x^2
f(x+d)=1/(x+d)^2
slope= lim as d>o of 1/d (1/( x+d)^2-1/x^2)=
= lim 1/d* (x^2-x^2-2dx-x^2)/(x^2*(x+d)^2) =lim (-2dx)/(x^2(x+d)^2)=-2/x^3
f(x+d)=1/(x+d)^2
slope= lim as d>o of 1/d (1/( x+d)^2-1/x^2)=
= lim 1/d* (x^2-x^2-2dx-x^2)/(x^2*(x+d)^2) =lim (-2dx)/(x^2(x+d)^2)=-2/x^3