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Lecture 4 3 June Corey Irving • Let's go back to Section 1.3 just for a moment: • The correct stretching and shrinking rules are: y= f ( cx) SHRINK y= f ( x) horizontally by factor of c x STRETCH y= f ( x) horizontally by factor of c y= f ( ) c • I had the words SHRINK and STRETCH reversed in Lecture 2. • Let's go back and do the examples in the notes that this error effected: ◦ Lecture 2, Section 1.3, example 1 part 6: x 2 y=( ) 2 • This was the second to last example we did in yesterday's notes: ◦ Graph the following: x f (x )=∣ln ( )∣ 2 1 Section 2.1: The Tangent and Velocity Problems • Section 2.1 Suggested Exercises: 1-9 • The tangent line to a curve at a point, P, on the curve is the line that most closely approximates the curve near P. Near P, this tangent line will hit the curve at the point P, but at no other points. • A secant line of a curve is formed by choosing two points on the curve and drawing the line determined by those two points. • How do we construct a tangent line given a graph and a point on it? The Velocity Problem: Example 2. The table shows distance d from Milner that I was at after time t on my walk to Blocker from Milner this morning. (Average adult walking speed: 3mi/hr= 264 ft/sec.) Time t (minutes) 1 2 3 4 Distance d (feet) 300 550 700 1000 1. What is the slope of the secant line through the points where t=1,2 t=2,3 2 t=3,4 2. What is the interpretation of these slopes? 3. Describe how we can find the instantaneous velocity at the point where t=3. Section 2.2: The Limit of a Function • Section 2.2 Suggested Homework: 1-8, 9, 11, 12, 13-19 odd, 21-25 • Intuitively, the limit of a function, y= f ( x) , asks the following: as x gets closer and closer to some value a , what y value is y= f ( x) getting closer and closer too? • The notation for the above intuitive notion is lim x → a f ( x) • If y= f ( x) does not approach exactly one real number value as x approaches a, we say that the limit does not exist. (HW #1) • At first glance, it may seem like as x gets closer to a, y= f ( x) always gets closer to y= f ( a). • This is very often what happens, but not always. For instance, it may be that a is not in the domain of f.. Consider the following examples: Example 1. 1. We have a denominator. 3 2. Piecewise functions • Definition. (HW #2) The left-hand limit of f (x ) as x approaches a is the value that f (x ) approaches as x approaches a from the left, or equivalently, the value that f ( x ) as x increases towards a. The notation for this is lim x → a f ( x) − • Similarly, The right-hand limit of f (x ) as x approaches a is the value that f ( x ) approaches as x approaches a from the right, or equivalently, the value that f ( x ) as x decreases towards a. The notation for this is lim x → a f ( x) + lim x → a f ( x)= L ⇔ lim x → a f ( x) And − lim x → a f ( x) + Example 2. (HW #3-8, 13-16) For which values of a does a limit exist? { } 1 x⩽−2 1. f ( x )= x 3 −2< x⩽1 2− x x>1 4 2. With f as above, evaluate lim x →−2 f ( x ) + Section 2.3: Calculating Limits Using the Limit Laws • Section 2.3 Suggested Homework: 1, 2, 3-7 odd, 8-24, 33-39, 49 • Assume that c is a constant and the limits lim x → a f ( x) and lim x → a g ( x ) exist, then we have the following limit Laws: The Limit Laws lim x → a [ f ( x)+g ( x)]=lim x→ a f ( x )+lim x → a g (x ) lim x → a [ f ( x)−g (x )]=lim x→ a f (x )−lim x → a g (x ) lim x → a f ( x)=c lim x → a f ( x) lim x → a [ f ( x) g ( x)]=lim x →a f (x )lim x→ a g ( x) 5 f ( x) lim x → a f ( x) lim x → a = g (x ) lim x → a g ( x) Example 1. If lim x → a f ( x)=π and lim x → a g ( x )=π 2 , evaluate the following • lim x → a [ f ( x) g ( x)] π f (x) • lim x → a g ( x) Example 2. n • lim x → a [ f ( x)] x 2−9 3 • lim x →−3 ( ) x+3 6 Here are more properties of limits: lim x → a √ f ( x )=√ lim x → a f ( x) 1. 2. If f is a polynomial or rational function and a is in the domain of f , then lim x → a f ( x)= f ( a) Example 3. 2 x −2 • lim x → 4 2 x −9 Example 4. Evaluate the limit 4−x 1. lim x → 4 + ∣4−x∣ x 2 −9 2. lim x →−3 x 2+2x−3 7 x 2+ x−12 3. lim x → 3 x−3 4. Trick: Calculating Limits by Rationalizing: lim x → 3 √ x+6−x x 3−3x 2 ∣x∣ 5. lim x → 0 x • You do not have to learn the Squeeze Theorem! 8

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