Question

Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is

Answer 1

So the answer to your question is:
1.  critical value g'(x) = 0
(x^2 - 16)/(x - 2) = 0
(x - 4)(x + 4) = 0
x = 4 or x = -4
g or G critical values at x = 4 and x = -4

2. g''(x) = [(x - 2)(2x) - (x^2 - 16)]/(x - 2)^2 = [2x^2 - 4x - x^2 + 16]/(x - 2)^2 = (x^2 - 4x + 16)/(x - 2)^2
g''(4) = 4^2 - 4(4) + 16 = 16 - 16 + 16 = 16 => local minimum
g''(-4) = (-4)^2 - 4(-4) + 16 = 16 + 16 + 16 = 48 => local minimum

3.  g is not concave down at any intervals.

4. y - 4 = g'(3)(x - 3)
y - 4 = [((3)^2 - 16)/(3 - 2)](x - 3)
y - 4 = -7(x - 3)
y = -7x + 25

5. So I'm assuming the tangent line lies below the graph.

Hope this helps!

Answer 2

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.

$\bf g'(x)=\cfrac{x^2-16}{x-2}\\\\ -------------------------------\\\\ x-2=0\implies \boxed{x=2} \\\\\\ 0=\cfrac{x^2-16}{x-2}\implies 0=x^2-16\implies 16=x^2\implies \boxed{\pm 4 = x}$


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   $\bf g''(x)=\cfrac{x^2-4x+16}{(x-2)^2}$

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

$\bf y-4=-7(x-3)\implies y=-7x+25\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}$

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.