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Calculus BC Final Exam Study Guide Semester 1 (Calc AB whole year)
Limits Note: for this review sheet
( ) is the same as
:: thethe limit as x approaches “a” from the left → →lim lim− limit as x approaches “a” from the right → →“The big limit” limit” – If the limit equals the limit from the
Introduction
→
→
Limit exists if the function is continuous or has a hole at “a” Limit does not exist (DNE) if the function has a V.A., a jump, or an endpoint If either of the limits from the left or right are o ∞ then the limit DNE
+
left, and the value is a real, finite number, then the limit exists
Laws of Limits
( ) lim→ () then: → ± = li→m ± li→m lim →lim ∗lim→ → ∗ = li→m lim → lim → = lim→
→ ∗ = ∗ lim→ lim → = lim→
Suppose: lim x a
x a
x a
lim x a
x a
x a
x a
x a
x a
x a
x a
x a
x a
Steps to solving limits (stop when limit is found)
−
1. Evaluate function at = if real finite number then the limit is found (make sure to check that limit is same from left as it is from the right +) 0 a. If you get then it is a hole 0 2. See if the function reduces by adding fractions or factoring etc. then re-evaluate (go to step 1) 3. If there is a radical, multiply by the conjugate to rationalize numerator/denominator then evaluate (go to step 1) Determine horizontal asymptote for lim No limit for slant asymptote (top heavy), approaches ∞ o If powers equal then use coefficients o Make sure to expand first If bottom heavy then lim =0 o Be specific when you write the graphical significance, for example Odd V.A. at = 2 If the limit approaches ∞ ( ) you can say that the limit is ∞ even though it DNE
→∞
→∞
→∞
Definitions
− →
→
Continuity – Continuity – a function has no jumps, holes (removable discontinuities), etc. ( ) is continuous at = if lim ( ) = lim + ( ) = ( ) note: ( ) must exist o Removable discontinuity (hole) – a point in the function where x is undefined Zero on top and bottom o
Force a Function to be Continuous
Example: Find ( , ) so
− → − − → −
is continuous at
= 1. 2 +
=
6
lim 6 1
+
1=4
+2
4 +
1
lim+
6 + 1=4 1+ +2 =4
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>1 =1 <1
1
2
+
=1 2 =2
+2 =4
1. Show limits from right & left sides 2. Setup system of eq. and substitute x value
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Intermediate Value Theorem
If f is continuous on [ , ] and “ ” is any number between ( ) and ( ) then there exists an between and such that ( ) = .
To verify the I.V.T. and find the root guaranteed by it… 1. Set k = 0 if finding root (root is where y value is 0) 2. Find ( ) and ( ) 3. Since at = f is ___ and at = is ___ and is continuous then ___ and ___ at the root. To find the root set ( ) equal to 0 and solve
must be zero between
Tangent Lines/Derivatives with Limits Approximate slope of tangent line: .
=
.
+
.
2
Derivatives – function that finds the slope of tangent lines Differentiable (adj) – derivative exists Differentiate (verb) – to find derivative For a derivative to exist Continuity at = o is continuous o Secant line must approach from both sides to the tangent line o must exist and be defined , ,& If then is smooth =
′ ′ → − − − ′ → → − − ′ ′
Definition 1 (RDQ):
= lim
+
0 ( )
Definition 2: = lim = lim ( ) is the slope of the line tangent to at (3) is the slope of the tangent line to at RDQ
′
Definition 1 (LDQ): ( )
→ −−
= lim
0
=3
Write tangent eq.
′ − − ′ ’≥ ’ ’≤ ′ − … 1. 2. 3. 4.
Find derivative function using definition 1 or 2 Substitute the value into function Find = ( ) and substitute in Fill in ( 1 = 1)
( ) Graphs
If is increasing then 0 If is positive, then f is increasing o 0 If is decreasing then If is negative, then is decreasing o
Derivatives
1 = = General Power Rule: if then Multiply coefficient by exponent, subtract 1 from exponent o ( ) means take the derivative in respect to the variable Find the derivative, treat all other variables except as constants o Definition of : the number whose base for which the curve = passes through (0,1) If then 0 = 1 = o
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= (ln ) =
= ∗ ln log = ∗ hint: convert log → Product Rule Quotient Rule = ∙ = ′ = ′ ∙ + ∙′ ′ ′ = ′ ∙−∙ ′ ′ ′ Motion: = () = = () means the change in y with respect to x Rate ⇒ change in something with respect to time o
o
1
o
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1
o
1
ln
ln
ln
o
o
o
o
o
Trig Derivatives
→ − → − − − −− sin
0
lim
− −
Note: avg. velocity is a secant line to the ( ) graph NOT the
o
lim
2
cos
1
0
sin
=1
=0
= cos
cos
=
tan
= sec 2
cot
Note: all co- functions have negative derivatives
Inverse Trig Derivatives sin
1
cos
1
=
=
2
1
1
2
tan
1
cot
1
sec
csc
csc 2
=
− − −
1
1
sin
=
=
2
1
1+
= tan sec
=
cot csc
− − − − −
1
1+
−
2
sec
1
csc
1
1
=
2
1
1
=
2
1
Chain Rule
Work from outside in taking derivatives & multiply them together Parametric Composite Function →
∗
′ ′ ∗′ ′ ′ ∗′ ∗′
= =
=
=
=
=
=
=
Implicit Relationship
Find the derivatives of all parts (using quotient, product, chain rules) Make sure to add to the y-variable derivatives
Solve for
L’Hopitals’ Rule o o
∞
0
∞∞ ±
If limits at a point or ± are in the form 0 ± (indeterminate form) then Take derivative of numerator & denominator separately & try again
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( )
( )
Linear Approximations
∞∞ 0
±
0
±
′ lim → = →lim ′
Δ −
Unknown ( 2 , =
2
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2)
1
Known (
1,
1)
Dy is used to approximate Δy Error = | | If curve is concave down then overestimate (see picture); if concave up then underestimate o
−Δ
Example (font size is small because this takes up too much roo m otherwise)
Δ − ≈ − ∗ ∗ ∗ ∗ ∗ ∗ ∗ ≈ =
3
2 +1
Find Δ y and as x changes from 13 13.1 First we find Δy using a calculator: →
=
13.1
13
.0074
Then we find dy. Differentiate the function solving for dy (don’t forget the chain rule). 1
Since we know
= 13 and
= 2 +1 3 2 1 = 2 +1 3 2 3 2 1 = 2 3 2 +1 3
1
= . 1 = we can substitute into the above equation and solve for dy 10 2 1 2 1 2 1 2 1 = = = = .0074 2 2 = 3 3 30 30 9 270 135 27 2 +1 3
≈
If we were to approximate (13.1) without a calculator:
13.1
13 +
=
3
27 +
1 135
Derivative Graphs
Fermat’s Theorem
o
′
is an endpoint
If is differentiable and has a local maxima or minima at = , then =0 Extreme Value Theorem If is continuous on [ , ] then has an absolute maxmima ( ) and a absolute minimum ( ) o at values and in [ , ] (note: must be closed interval!) A graph is concave down if its tangents are above the graph Frownie-face (down = unhappy) o o
′ ′ ′ ′′
′ ′ ′
Finding maxes and mins based on info (given is zero or undefined) If changes sign from + to – at = , then ( ) is a min o If changes sign from – to + at = , then ( ) is a max o Critical Points: c is a critical number is undefined o =0 o
o o
< 0
is decreasing
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A graph is concave up if its tangents are below the graph Smiley-face (up=happy) o o
>0
is increasing Points of Inflection: where concavity changes There are points of inflection where = 0 because it changes sign o Mean Value Theorem (MVT) If is continuous over [ , ] and differentable over ( , ) then there exists at least one in ( , ) o such that it is equal to the slope between and . Basically, there’s a derivative value ( ) somewhere on the graph that equals the slope between o and if its continuous and differentiable o
′′
′
Integrals and Anti-derivatives
Field lines
Draw slope of the tangent line at each point (value of derivative)
Basics
+ == −2−4 +4 finding derivative ∫ 2−−44 finding anti-derivative/ integral (integrating) Memorized derivatives apply in reverse General power rule: ∫ = + 2
←
2
←
+1
+1
Trig Anti-Derivatives
∫ ∫ ∫ − ∫ − ∫ ∫ − ∫
∫ ∫ ∫
∫ − ∫ −− ∫
∫ − ∫ − −
cos
= sin
=
csc2
+
(csc ) = l n csc tan
sec 2
+
cot
= ln sec
+
2
1
1
1
2
= arcsin +
= arccos +
Other Anti-Derivatives =
ln
=
+
cot +
= ln sec + tan
+
+
Inverse Trig Anti-Derivatives 1
sec
= tan
∫ ∫ − 1
1+
2
2
= arccot +
1
1+
= arctan +
+
1st Fundamental Theorem of Calculus (FTC)
If f is continuous over [a,b] then
′ = − Now possible to evaluate definite integrals
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tan sec
= sec + =
cot
= ln sin
1
2
1
1 2
1
+
+
= arcsec + = arccsc +
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Approximation of Area Rectangles
≈ ⋯− ≈ ⋯− ≈ ⋯ −
Right sums area 1
+
2
+
1
note: this uses the right side of the partitions
+
Left sums area [
+
1
+
2
+
1
]
Midpoint sums area 0+ 1
2
+
1+ 2
2
1+
+
2
Trapezoids
− partitions =
=
+ 2( + 2( + ⋯ 2 − +
Trap area ≈ 2 [
1
2
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1
2
1
+
2
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Volume by Integration Disk Method
Washer Method
=
Shell Method
= −
2
2
= 2∗∗ Where = or and
2
∥
rectangles to axis of rotation
Solids with known cross sections
Watch for the radius doesn’t go to an axis:
= ∗ Where = or
− =4
dy
Arc Length
=4
+ 1 = + = = 1 + Average Value (height) of a Function ∫ − = = (−) 2
2
Engineering Applications
∗ ∗ =
Hooke’s law: = where x is the displacement from the natural length of the spring
=
=
Were
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2
2
is the force and
is the distance
