Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is
g′(x) = quotient of quantity x squared minus sixteen and x minus two with x ≠ 2.
find all values of x where the graph of g has a critical value.
for each critical value, state whether the graph of g has a local maximum, local minimum, or neither. you must justify your answers with a complete sentence.
on what intervals is the graph of g concave down? justify your answer.
write an equation for the tangent line to the graph of g at the point where x = 3.
does this tangent line lie above or below the graph at this point? justify your answer.

1)

well, this part is simple, since the derivative is given, so, you get critical points from the denominator being 0, or the derivative itself being 0.  When the denominator is set to 0, you get a critical value, however, this is "cusp", namely, as I mentioned before, for a graph to be differentiable, it has to be a smooth transition, but cusps are just an abrupt edge or an asymptote border, however abrupt they may be, they are indeed, a spike up or down in the graph, and thus a critical point.



2)

this, like the one yesterday in the optimization posting, is just a matter of doing a first-derivative test at those points, check the picture below.

3)

now the second derivative is   

if you zero out the derivative itself, the numerator's quadratic, doesn't yield any inflection points, just imaginary values, meaning there aren't any.

now, the bottom does, gives us a 2, however, judging from the critical point of 2 from the first derivative, chances are, that's an asymptote border.

so... if you run a second-derivative test on those regions, say check g''(0) and g''(3), you'll notice, you get positive on both ends, so is concave up before and after the 2, so is always concave up.

quick note:  bear in mind that, on doing either first or second derivative tests, the value that you get is not very meaningful, what matters is the sign, is it positive or negative?

4)

we know g(3) = 4, is just another way of saying x = 3, y = 4, now, what's the slope? well, let's use the first derivative.  Anyway, if you do a g'(3), you'd get a -7 for the slope at that x  = 3. thus



5)

well, recall the graph from section 2)
recall on what region the 3 is at, is coming from the top and going downwards, reason why is a negative slope anyway, so is above.