The line defined by the equation 2y+3=-(2/3)(x-3) is tangent to the graph of g(x) at x=-3. What is the value of the limit as x approaches -3 of [g(x)-g(-3)]over(x+3)?

Question

The line defined by the equation 2y+3=-(2/3)(x-3) is tangent to the graph of g(x) at x=-3.  What is the value of the limit as x approaches -3 of [g(x)-g(-3)]over(x+3)?

larseighner · Answer

What is the slope of the tangent line?

precal · Answer

ok I would have to use the given equation to determine this, give me a moment

precal · Answer

slope of the tangent line is -1/3

larseighner · Answer

okay so what does the limit you mention mean?

precal · Answer

no clue

precal · Answer

is it one of the alternate definitions of a derivative?

precal · Answer

I meant  g ' (-3) perhaps?

larseighner · Answer

I think so.  What is the definition of g'(x) at -3?

precal · Answer

is it -1/3?

precal · Answer

limit approaches a of [f(x)-f(a)]over(x-a)
is it this one?

larseighner · Answer

Okay the point here (well some of them) is

$ \displaystyle {{\Delta y} \over {\Delta x}} $

becomes the derivative however you take the limit that make $\Delta x$ approach zero

larseighner · Answer

and the derivative is the slope of the tangent line.  Usually you are given the limit to solve to find the slope of the tangent.  This is a switcharoo. They give you the tangent line, you can figure the slope. Your job was to recognize that the limit as they put is the derivative at the tangen point.

precal · Answer

thanks, I did not recognize this alternate definition of the derivative using the limit.  Thank you