1. 4.5 An Algorithm for Curve Sketching
We learned a bunch of “pieces” to curve sketching, now let’s put it all together …
STEP 1: Determine the DOMAIN of the function and all discontinuities (𝑓(𝑥) is undefined)
Polynomials are continuous and differentiable for ∀𝑥∈ℝ.
Some functions do have restrictions on their domains (i.e. 𝑓 𝑥 = 𝑥 is only defined for𝑥≥0).
Discontinuities can be infinite, removable or jump.
Infinite: 𝑓 𝑥 = 𝑥+1 𝑥−3 , 𝑓 3 is undefined (den=0 but num≠0) ⇒ A VERTICAL ASYMPTOTE
→ Don’t forget to check whether 𝑓(𝑥) approaches ±∞ at these points.
Removeable: 𝑓 𝑥 = (𝑥+1)(𝑥−3) 𝑥−3 , 𝑓 3 is undefined (den=0 and num=0) ⇒ A HOLE
Jump: think piecewise functions

2. 4.5 An Algorithm for Curve Sketching
STEP 2: Determine any 𝑥 and 𝑦 intercept(s).
STEP 3: Determine any horizontal or oblique asymptotes.
Horizontal Asymptotes & End Behaviour
Explore lim 𝑥→+∞ 𝑓(𝑥) and lim 𝑥→−∞ 𝑓(𝑥) to determine if there are any horizontal asymptotes.
→ Don’t forget to check whether 𝑓(𝑥) approaches the asymptote from above or below in each direction.
Oblique Asymptotes
The occur for all rational functions in which the degree of the numerator is exactly one greater than that of the denominator.
→ Again check whether 𝑓(𝑥) approaches the asymptote from above or below in each direction.

3. 4.5 An Algorithm for Curve Sketching
STEP 4: The First Derivative Test
1. Find all the CRITICAL POINTS, 𝒄, where 𝒇 ′ 𝒄 =𝟎 or 𝒇 ′ 𝒄 is undefined for 𝒄 in the domain of 𝒇.
It is important not to forget the points where 𝑓′(𝑐) because these can produce local max/min.
2. Classify all critical points as local maximum, local mininum or neither
→ set up an interval chart and test a value in each interval to determine whether 𝑓′(𝑥) is positive or negative and use this to classify .

4. 4.5 An Algorithm for Curve Sketching
STEP 5: The Second Derivative Test
1. Find all the points, 𝒙, where 𝒇 " 𝒙 =𝟎 or 𝒇 " 𝒙 does not exist.
NOTE: this will automatically include points where 𝑓(𝑥) or 𝑓 ′ 𝑥 is undefined.
2. Determine whether these are points of inflection.
→ set up an interval chart and test a value in each interval to determine whether 𝑓"(𝑥) is positive or negative and use this to classify points of inflection.
STEP 6: Graph the Function

5. Sketch the following function
Practice:
Sketch the following function
𝑓 𝑥 = 3𝑥 4 −8 𝑥 3 +6 𝑥 2
Step 1: 𝑓(𝑥) is a polynomial function ⇒ 𝐷 𝑓 =ℝ.
Step 2: 𝑓(𝑥) is a polynomial function ⇒ 𝑓 𝑥 is continuous for ∀𝑥∈ℝ.
Step 3: When 𝑓 0 =0⇒the 𝑦−intercept occurs at 0,0 .
𝑓 𝑥 =0⇒0= 3𝑥 4 −8 𝑥 3 +6 𝑥 2
⇒0= 𝑥 2 (3𝑥 2 −8𝑥+6)
𝑏 2 −4𝑎𝑐=64−4 18 =−8<0
⇒3 𝑥 2 −8𝑥+6 has no real roots
⇒ there is only one 𝑥−intercept at (0,0)
Step 4: 𝑓(𝑥) is a polynomial function ⇒ there is no horizontal or oblique asymptote.

6. Sketch the following function
Practice:
Sketch the following function
𝑓 𝑥 = 3𝑥 4 −8 𝑥 3 +6 𝑥 2
Step 5: 𝑓(𝑥) is a polynomial function ⇒ 𝑓′(𝑥) is defined for ∀𝑥∈ℝ.
𝑓 ′ 𝑥 =12 𝑥 3 −24 𝑥 2 +12𝑥
𝑓′ 𝑥 =0⇒0=12 𝑥 3 −24 𝑥 2 +12𝑥
⇒0= 12𝑥 (𝑥 2 −2𝑥+1)
⇒0= 12𝑥 (𝑥−1) 2
⇒𝑥=0 and 𝑥=1 are critical points of 𝑓.
Note: 𝑓 1 =1
Interval
𝒙<𝟎
𝟎<𝒙<𝟏
𝒙>𝟏
Sign of 𝑓′(𝑥)
− + =−
+ + =+
𝑓(𝑥)
decreasing
increasing
∴ there is a local minimum at (0, 0).

7. Sketch the following function
Practice:
Sketch the following function
𝑓 𝑥 = 3𝑥 4 −8 𝑥 3 +6 𝑥 2
𝑓 ′ 𝑥 =12 𝑥 3 −24 𝑥 2 +12𝑥
Step 6: 𝑓(𝑥) is a polynomial function ⇒ 𝑓"(𝑥) is defined for ∀𝑥∈ℝ.
𝑓" 𝑥 =36 𝑥 2 −48𝑥+12
𝑓" 𝑥 =0⇒0=36 𝑥 2 −48𝑥+12
⇒0=3 𝑥 2 −4𝑥+1
⇒0=(𝑥−1)(3𝑥−1)
⇒𝑥=1 and 𝑥= 1 3 .
Interval
𝒙< 𝟏 𝟑
𝟏 𝟑 <𝒙<𝟏
𝒙>𝟏
Sign of 𝑓"(𝑥)
− − =+
− + =−
+ + =+
𝑓(𝑥)
concave up
concave down
∴ there are points of inflection at , and (1,1).

8. In summary …
QUESTIONS: p #1, 2, 3a, 4, 5, 6