Question

Calculus Help

Answer 1

they give you the first derivative of g. g ' (x)

maybe start off by figuring out the function g(x) and thsecond derivative g ''(x)

Answer 2

The key critical values of a function g(x) occur where g'(x)=0 (Stationary Points), or g'(x) is undefined.

Answer 3

I'm not sure how to "un-derive" but the second derivative I got was f''(x)=\[x^2-4x+16\]

Answer 4

well just g '' don't think you need g

Answer 5

oops i forgot the denominator. x^2-4x+16/(x-2)^2

Answer 6

for number 1, I set the first derivative to 0 and DNE. and I got -4,2,4

Answer 7

those should be your critical values. Concave up/down is determined by the sign of g''(x) at the key point. Simply evaluate g''(-4), g''(2), g''(4). If g''(x)>0, the graph is concave up there. If g''(x)<0, the graph is concave down there.

Answer 8

nevermind, sorry didn't read the question, you don't need that

Answer 9

I thought I had to set the second derivative to 0 and DNE to find the inflection points. And those determine concave up or down? Thats number 3

Answer 10

g''(x)=0 and g''(x) DNE gives inflection points, yes. If g''(x)<0 for values in the interval between inflection points, then the graph of g is concave DOWN for that interval. Similarly, id g''(x)>0 for values in the interval between inflection points, the the graph of g is concave UP on that interval.

Answer 11

pick a value inside each interval, test it in g '(x), to see if you get + or -, inc or dec

Answer 12

I had a problem setting the second derivative = to 0 and DNE. I couldn't get any points

Answer 13

@DanJS I got decreasing, increasing, decreasing, decreasing

Answer 14

ok, the points are the critical points for #1

#2, say that for each interval (-inf,-4), (-4,-2)...

Answer 15

wait, x=5 gives + number, that is increasing

Answer 16

I made the chart



youre right about the last one