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Part 1: Critical Points & Increasing/Decreasing
Applications of Derivatives
Part 1 of 7 โ Critical Points & Increasing/Decreasing
Critical Points
A critical point of f occurs where:
- fโฒ(x)=0, or
- fโฒ(x) is undefined (but f(x) exists)
First Derivative Test for Increasing/Decreasing
| fโฒ(x) | Behavior of f |
|---|
| f |
Worked Example
Find where f(x)=x3โ3x+1 is increasing and decreasing.
fโฒ(x)=3x2โ3=3(x
Critical points: x=โ1 and x=1.
| Interval | fโฒ(x) | Behavior |
|---|
| (โโ,โ1) | |
First Derivative Test for Local Extrema
At a critical point c:
| Sign change of fโฒ | Conclusion |
|---|
| +โโ | Local maximum at |
Classify Critical Points ๐ฏ
Key Takeaways โ Part 1
- Critical points occur where fโฒ=0 or fโฒ is undefined
- โ increasing, โ decreasing
Part 2: Second Derivative & Concavity
Applications of Derivatives
Part 2 of 7 โ Second Derivative & Concavity
Concavity
| fโฒโฒ(x) | Concavity |
|---|
| f |
Part 3: Absolute (Global) Extrema
Applications of Derivatives
Part 3 of 7 โ Absolute (Global) Extrema
Extreme Value Theorem (EVT)
If f is continuous on a closed interval [a,b], then f attains an absolute maximum and absolute minimum on [a.
Part 4: Curve Sketching
Applications of Derivatives
Part 4 of 7 โ Curve Sketching
Complete Curve Sketching Procedure
- Domain of f
- Intercepts: set y=0 (x-intercepts) and x=0 (y-intercept)
- Symmetry: even () or odd ()
Part 5: Mean Value Theorem
Applications of Derivatives
Part 5 of 7 โ Mean Value Theorem
Statement (MVT)
If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c in such that:
Part 6: Related Rates (Mini-Review)
Applications of Derivatives
Part 6 of 7 โ Related Rates (Mini-Review)
Related Rates Strategy
- Draw a picture and label variables
- Write an equation relating the variables
- Differentiate both sides with respect to time t
- Substitute known values and solve
Related Rates & Applications ๐ฏ
Key Takeaways โ Part 6
- Related rates: differentiate with respect to t using Chain Rule
- Always identify what you're given and what you're finding
- Don't substitute values until AFTER differentiating
Part 7: Comprehensive Assessment
Applications of Derivatives โ Review
Part 7 of 7 โ Comprehensive Assessment
Applications of Derivatives โ Complete! โ
You have mastered:
- โ
Critical points and increasing/decreasing analysis
- โ
Second derivative test and concavity
- โ
Absolute extrema on closed intervals
- โ
Curve sketching
- โ
Mean Value Theorem
- โ
Related rates applications
โฒ
(
x
)
>
0
| fโฒ(x)<0 | f is decreasing |
+
1)(xโ
1)
+
| (โ1,1) | โ | Decreasing |
| (1,โ) | + | Increasing |
x=c
| โโ+ | Local minimum at x=c |
| No sign change | Neither (e.g., inflection point) |
Example (continued)
For f(x)=x3โ3x+1:
- At x=โ1: fโฒ changes from + to โ โ local max at f(โ1)=3
- At x=1: fโฒ changes from โ to + โ local min at
fโฒ
>
0
First Derivative Test: sign change determines max/min/neitherโฒโฒ
(
x
)
>
0
| Concave up (opens upward, "cup") |
| fโฒโฒ(x)<0 | Concave down (opens downward, "cap") |
Inflection Points
An inflection point is where concavity changes. This occurs where fโฒโฒ=0 or fโฒโฒ is undefined, AND fโฒโฒ actually changes sign.
Second Derivative Test
At a critical point where fโฒ(c)=0:
- If fโฒโฒ(c)>0: local minimum (concave up)
- If fโฒโฒ(c)<0: local maximum (concave down)
- If fโฒโฒ(c)=0: inconclusive (use First Derivative Test)
Worked Example
f(x)=x3โ6x2+9x+1
fโฒ(x)=3x2โ12x+9=3(xโ1)(xโ3). Critical points: x=1,3.
fโฒโฒ(x)=6xโ12.
- fโฒโฒ(1)=โ6<0 โ local max at x=1
- fโฒโฒ(3)=6>0 โ local min at x=3
- fโฒโฒ(x)=0 at x=2 โ inflection point
Second Derivative Analysis ๐ฏ
Key Takeaways โ Part 2
- fโฒโฒ>0: concave up; fโฒโฒ<0: concave down
- Inflection points: where fโฒโฒ changes sign
- Second Derivative Test: faster than First Derivative Test when fโฒโฒ(c)๎ =0
,
b
]
Candidates Test (Closed Interval Method)
- Find all critical points in (a,b)
- Evaluate f at each critical point AND at the endpoints a and b
- The largest value is the absolute max; the smallest is the absolute min
Worked Example
Find the absolute extrema of f(x)=x3โ3x+1 on [โ2,2].
fโฒ(x)=3x2โ3=0 โ x=ยฑ1 (both in the interval).
| x | f(x) |
|---|
| โ2 | โ8+6+1=โ1 |
| โ1 | โ1+3+1=3 |
| 1 | 1โ3+1=โ1 |
| 2 | 8โ6+1=3 |
Absolute max = 3 (at x=โ1 and x=2). Absolute min = โ1 (at x=โ2 and x=1).
Key Takeaways โ Part 3
- EVT guarantees extrema on closed intervals for continuous functions
- Candidates Test: evaluate at critical points AND endpoints
- Compare all values to find the absolute max and min
f(โx)=f(x)
f(โx)=โf(x) First derivative: critical points, increasing/decreasing, local extremaSecond derivative: concavity, inflection pointsEnd behavior: limxโยฑโโf(x) Asymptotes (if any)Quick Example
f(x)=x4โ4x3
Curve Sketching from Derivatives ๐ฏ
Given fโฒ(x)=(xโ1)2(xโ4):
Key Takeaways โ Part 4
- A systematic approach using both fโฒ and fโฒโฒ gives a complete picture
- From a graph of fโฒ, you can deduce everything about f's shape
- This is a frequent AP free-response topic
(a,b)
fโฒ(c)=bโaf(b)โf(a)โ
Geometric meaning: There's a point where the tangent line is parallel to the secant line through (a,f(a)) and (b,f(b)).
Worked Example
f(x)=x3 on [0,2].
Average rate of change: 2โ0f(2)โf(0)โ=28โ0โ=4.
Find c: fโฒ(c)=3c2=4 โ c=3โ2โ=.
Since cโ(0,2), MVT is confirmed.
Key Takeaways โ Part 5
- MVT: there's a point where instantaneous rate = average rate
- Must check: f continuous on [a,b], differentiable on (a,b)
- MVT is used heavily in AP justifications and proofs
f(1)=โ1
x=3
f(3)=81โ108=โ27
โ
2)
- Inflection at x=0 and x=2
- Concave up on (โโ,0) and (2,โ)
- Concave down on (0,2)
โ
โ
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