1. Let’s get our brains back on track …
What is our goal for this unit?
To sketch curves of functions … woohoo!
What type of functions?
Polynomial and Rational
What do we know so far?
The domain of a function, including any discontinuities (jump, holes/removeable, vertical asymptotes, or if f(x) has any end point due to restrictions on it’s domain).
End behaviours, horizontal and oblique asymptotes.
THE FIRST DERIVATIVE TEST - Intervals of increase/decrease, max/min points.
So … what’s left?

2. 4.4 Concavity and Points of Inflection

3. 4.4 Concavity and Points of Inflection
Let’s re-examine the graph of y=𝑓 𝑥 = 𝑥 3 and now look at concavity. In particular, let’s look at what’s happening to the slope in the two intervals, 𝑥<0 and 𝑥>0.
We know that the slope is positive on both of these intervals (i.e. 𝑓 ′ 𝑥 >0). But what we are interested in now is how the slope is changing as 𝑥 moves from left to right on these intervals. Do we have any “tool/equation” that tells us about how the slope of a graph changes with respect to 𝑥?
Recall that, simply put,
𝑓 ′ 𝑥 = ∆𝑓(𝑥) ∆𝑥
𝑓 " 𝑥 = ∆𝑓′(𝑥) ∆𝑥
and

4. 4.4 Concavity and Points of Inflection
So, let’s look at the graph of 𝑦 ′ = 𝑓 ′ 𝑥 =3 𝑥 2 , in particular the SLOPE/RATE OF CHANGE of this graph.
Note that the slope of 𝑓′(𝑥) is decreasing throughout the interval when 𝑥<0 (i.e. 𝑓"(𝑥)<0) and is increasing throughout the interval when 𝑥>0 (i.e. 𝑓"(𝑥)>0).

5. DEFINITIONS
Suppose a function, 𝑓, is twice differentiable on the interval 𝑎<𝑥<𝑏.
A graph, 𝑦=𝑓(𝑥), is concave upward on the interval 𝑎<𝑥<𝑏 if 𝑓"(𝑥)>0 for 𝑎<𝑥<𝑏. Graphically, the curve 𝑦=𝑓(𝑥) lies above the tangent line at each point in the interval 𝑎<𝑥<𝑏.
A graph, 𝑦=𝑓(𝑥), is concave downward on the interval 𝑎<𝑥<𝑏 if 𝑓"(𝑥)<0 for 𝑎<𝑥<𝑏. Graphically, the curve 𝑦=𝑓(𝑥) lies below the tangent line at each point in the interval 𝑎<𝑥<𝑏.
Assuming that the function is continuous, a point of inflection separates an interval where the curve is concave upward from an interval where the curve is concave downward BUT ...
NOT ALL values of 𝑥 having 𝑓"(𝑥)=0 will be points of inflection.

6. TO CHECK WE MUST SET UP INTERVALS OF CONCAVITY
To confirm POINTS of inflection we must check if 𝑓"(𝑥) changes its sign at that value (similar to when we check for extreme values using the first derivative).
We must also check this for any points where 𝑓 ′ 𝑥 or 𝑓"(𝑥) is undefined, since they may also separate the intervals of concavity.
What does this look like? ….

7. EXAMPLE 1: VERTICAL TANGENT (𝒇′(𝒙) does not exist)
Consider 𝑓 𝑥 = 3 𝑥 .
𝑓′ 𝑥 = 1 3 𝑥 2 3
𝑓′ 0 does not exist ⇒𝑓"(𝑥) cannot exist.
However, there is a point of inflection at 𝑥=0.

8. EXAMPLE 2: 𝒇 ′ 𝒄 exits, 𝒇"(𝒄) does not exist and (𝒄,𝒇 𝒄 ) is NOT an inflection point.
Consider 𝑓 𝑥 = 𝑥
𝑓′(𝑥)= 4 3 𝑥 and 𝑓"(𝑥)= 4 9 𝑥 2 3
⇒𝑓 ′ 0 =0 and 𝑓"(0) does not exit.
We see from the graph that (0,0) is not an inflection point.

9. EXAMPLE 3: 𝒇 ′ 𝒄 exits, 𝒇"(𝒄) does not exist and (𝒄,𝒇 𝒄 ) IS an inflection point
Consider 𝑓 𝑥 = 𝑥
𝑓′(𝑥)= 5 3 𝑥 and 𝑓"(𝑥)= 10 9 𝑥 1 3
⇒𝑓 ′ 0 =0 and 𝑓"(0) does not exit.
Very similar results; however, we see from the graph that (0,0) is an inflection point.

10. THE SECOND DERIVATIVE TEST
If the second derivative 𝑓"(𝑥) is continuous on an open interval that contains 𝑥=𝑐 and 𝑓 ′ 𝑐 =0, then
if 𝑓"(𝑐)>0, then 𝑓(𝑥) is concave upward at 𝑥=𝑐⇒𝑓(𝑥) has a local minimum at 𝑐, and
if 𝑓"(𝑐)<0, then 𝑓(𝑥) is concave downward at 𝑥=𝑐⇒𝑓(𝑥) has a local maximum at 𝑐
NOTE:
The second derivative test does not address the case where 𝑓"(𝑥)=0 or where 𝑓"(𝑥) does not exist. In these cases, we must use these values as boundary points to set up a table of intervals of concavity and test 𝑓"(𝑥) in each interval to determine if these are points of inflection. We must use the first derivative and our intervals of increase/decrease to make conclusions about local max/min.

11. Find the intervals of concavity for the following functions.
Practice:
Find the intervals of concavity for the following functions.
(a) 𝑓(𝑥)=𝑥 𝑥+1
Domain = 𝑥∈ℝ 𝑥≥−1 and 𝑓 𝑥 is continuous over its domain.
𝑓 ′ 𝑥 = 3𝑥+2 2 𝑥
𝑓 ′ −1 and 𝑓"(−1) do not exist and 𝑥=−1 is in the domain of 𝑓 𝑥 .
𝑓" 𝑥 = 3𝑥+4 4 𝑥
𝑓" 𝑥 =0⇒3𝑥+4=0⇒𝑥= −4 3 which is not in the domain of 𝑓(𝑥)
Interval
𝒙>−𝟏
Sign of 𝑓"(𝑥) +
𝑓(𝑥)
concave up

12. Find the intervals of concavity for the following functions.
Practice:
Find the intervals of concavity for the following functions.
(a) 𝑓 𝑥 = 3𝑥 4 −8 𝑥 3 +6 𝑥 2
𝑓(𝑥) is a polynomial function ⇒ it is continuous and differentiable for ∀𝑥∈ℝ.
𝑓 ′ 𝑥 =12 𝑥 3 −24 𝑥 2 +12𝑥
𝑓" 𝑥 =36 𝑥 2 −48𝑥+12
0=36 𝑥 2 −48𝑥+12
0=3 𝑥 2 −4𝑥+1
Interval
𝒙< 𝟏 𝟑
𝟏 𝟑 <𝒙<𝟏
𝒙>𝟏
Sign of 𝑓"(𝑥)
− − =+
− + =−
+ + =+
𝑓(𝑥)
concave up
concave down
0=(𝑥−1)(3𝑥−1)
𝑥= 1 3 , 1
∴𝑥= 1 3 and 𝑥=1 are both points of inflection .

13. In summary …
QUESTIONS: p #1, 2, 9, 10