Blog 11?

Alrightttt, so this week we went back to chapter three, and we are on section 3.4 which is continuity. It is verrry easy surprisingly! The whole thing is just taking first derivatives and second derivatives, while using the first derivative test. **Doing this will help you determine whether the graph is concave up or down, and if you do all 7 steps, it will help you to find the points of inflection.
Here are the 7 steps:
1. Take 1st derivative=0 Solve.
2. Take 2nd derivative=0 Solve.
3. Differentiable (Remember what makes a graph differentiable)
4. Set up intervals
5. Pick number’s on an interval, and plug into **2nd derivative
6. If the numbers positive the graph is concave up, if negative concave down
7. If there is a change it is a point of inflection

Example: Determine the open intervals on which the graph is concave up or down
**Notice this problem isn’t looking for points of inflection, so you wouldn’t have to work step 7.
-x^3+6x^2-9x-1
F’(x)=-3^2+12x-9
F”(x)=-6x+12
X=2
(-infinity,2) u (2,infinity)
F(1)=-6(1)+12=6
F(3)=-6(3)+12=-6
(-infinity,2) is concave up (infinity, 2) is concave down

Example 2:
X^2+1/x^2-1
F’(x)=-4x/x^2-1^2
F”(x)=4(-1+4x^2)/(x^2-1)^2
(-infinity,-1),(1, infinity) Concave up; (-1,1) Concave down
**With fractions, if you use quotient rule to solve for second derivative numbers will always cancel out from the top and the bottom making it easier to solve.

Now for something I don’t understand! I don’t understand at all how to find points of inflection, I know what it is, I just don’t know how to find it. I’m not sure if you have to graph the function then look at it? Or your just suppose to know where the graph changes?