Study Guides for MATH 100 at University of British Columbia - Okanagan (UBC OKANAGAN)


UBC OKANAGAN, UBC OKMATH 100Wayne BroughtonFall

MATH 100 Lecture Notes - Lecture 25: Derivative Test, Second Derivative, Inflection

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30 Nov 2018
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UBC OKANAGAN, UBC OKMATH 100Wayne BroughtonFall

MATH 100 Lecture Notes - Lecture 24: Maxima And Minima

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28 Nov 2018
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UBC OKANAGAN, UBC OKMATH 100Wayne BroughtonFall

MATH 100 Lecture 23: Maxima & minima

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22 Nov 2018
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UBC OKANAGAN, UBC OKMATH 100Wayne BroughtonFall

MATH 100 Lecture 22: linear approximation and differentiation

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21 Nov 2018
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UBC OKANAGAN, UBC OKMATH 100Wayne BroughtonFall

MATH 100 Lecture Notes - Lecture 19: Inverse Trigonometric Functions, Inverse Function, Logarithm

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8 Nov 2018
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Math 100 lecture 19 derivative of inverse functions and implicit functions. = sin is not one-to-one, so an inverse function cannot be found. The line c
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UBC OKANAGAN, UBC OKMATH 100Wayne BroughtonFall

MATH 100 Lecture Notes - Lecture 18: Inverse Function, Binary Logarithm, Logarithm

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8 Nov 2018
0
Math 100 lecture 8 inverse functions. Partial proof of the chain rule: =() = () , =() =( + ) (), = , 0. = . , 0 lim 0 = lim 0 . lim 0 the general formu
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UBC OKANAGAN, UBC OKMATH 100Wayne BroughtonFall

MATH 100 Study Guide - Fall 2018, Comprehensive Midterm Notes - Thomas Robert Shannon Broughton, Trigonometric Functions, Graph Of A Function

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5 Nov 2018
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Math 100 lecture 1 diagnostic test & solutions: x = 2 is a root for the polynomial x3 x2 8x + 12. Find all of the roots of this polynomial by factoring
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UBC OKANAGAN, UBC OKMATH 100AllWinter

MATH 100 Exam Solutions Winter 2018: Quotient Rule, Chain Rule, Squeeze Theorem

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16 Oct 2018
0
From the general definition of derivative, the derivative of (cid:1858)(cid:4666)(cid:1876)(cid:4667) at (cid:1853) is given by (cid:1858)(cid:4593)(ci
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UBC OKANAGAN, UBC OKMATH 100AllWinter

MATH 100 Midterm: MATH100 Midterm 2009 Winter Solutions

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10 Oct 2018
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The limit at c+ and c- will different values as the function is defined differently at c and at value less than c. Appling limit we get: (cid:958)(cid:
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UBC OKANAGAN, UBC OKMATH 100AllWinter

MATH 100 Exam Solutions Winter 2018: Squeeze Theorem, Asymptote, Negative Number

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10 Oct 2018
0
Factoring out both numerator and denominator we get: the function (cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:3404)(cid:3)(cid:3051)(cid:3118)(cid:3)(
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UBC OKANAGAN, UBC OKMATH 100AllWinter

MATH 100 Exam Solutions Winter 2018: Product Rule, Indeterminate Form, Squeeze Theorem

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10 Oct 2018
0
From the general definition of derivative, the derivative of (cid:1858)(cid:4666)(cid:1876)(cid:4667) at (cid:1853) is given by (cid:1858)(cid:4593)(ci
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UBC OKANAGAN, UBC OKMATH 100AllWinter

MATH 100 Exam Solutions Winter 2018: Hypotenuse, Power Rule, Product Rule

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10 Oct 2018
0
[6 marks] using only the de nition of the derivative, and not the rules, nd f(cid:2) (x) for the. 1. function f (x) = x2 + 1. Solution: f(cid:2) f(cid:
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