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**aBbN749** Let f & g be functions that are differentiable for all real values of x with

g(x) = (f(x))/x. If y=x-3 is an equation of the tangent line to the graph of f at x=1, what is the equation of the tangent line to the graph of g at x=1.

== So f '(1) = 1 = the slope of y = x - 3, and since g '(x) = [xf '(x) - f(x)]/x^2 then g '(1) = 1 - f(1).

But at x = 1 the line y = x - 3 gets the value y(1) = 1 - 3 = -2 ==>

the line passes through (1, -2) and this point is the tangency point ==> (1, -2) belongs to the graph of f ==> f(1) = -2, so finally:

g '(1) = 1 - f(1) = 1 -(-2) = 3

and

Let f(t)=1/t for t>0. For what value of t is f '(t) equal to the average rate of change of f on [a,b].

thanks