f(x) continuous, lim[x->3] (f(x)+8)/(x-3)=12; find eqn of tangent line

kernel6903

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Does anybody know how to solve this?



Assume that f (x) is everywhere continuous and it is given to you that:

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 3}\, \)\(\displaystyle \dfrac{f(x)\, +\, 8}{x\, -\, 3}\, =\, 12\)

It follows that y = _________ is the equation of the tangent line to y = f (x) at the point ( ____, ____ ).
 
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Does anybody know how to solve this?
Yes.

Assume that f (x) is everywhere continuous and it is given to you that:

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 3}\, \)\(\displaystyle \dfrac{f(x)\, +\, 8}{x\, -\, 3}\, =\, 12\)

It follows that y = _________ is the equation of the tangent line to y = f (x) at the point ( ____, ____ ).
Hint: What is the limit-based definition of the derivative L of f(x) at the value x = a (the definition that does not use "h")? Comparing this definition with the given limit expression, what is the value of L in this case? What must be the value of "a"? What must be the value of y = f(a)? ;)
 
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