3 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 3 The Derivative.

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3 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 3 The Derivative

3 - 2 © 2012 Pearson Education, Inc.. All rights reserved. Section 3.1 Limits

3 - 3 © 2012 Pearson Education, Inc.. All rights reserved.

3 - 4 © 2012 Pearson Education, Inc.. All rights reserved. Figure 2

3 - 5 © 2012 Pearson Education, Inc.. All rights reserved. Notation *from Spivak’s Calculus

3 - 6 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn Solution: since The numerator also approaches 0 as x approaches −3, and 0/0 is meaningless. For x ≠ − 3 we can, however, simplify the function by rewriting the fraction as Now

3 - 7 © 2012 Pearson Education, Inc.. All rights reserved. Left and Right

3 - 8 © 2012 Pearson Education, Inc.. All rights reserved. Left and Right  What can you say about lim f(x) as x  10 if  lim f(x) as x  10 - (from the left) is 5  lim f(x) as x  10 + (from the right) is 5 ?

3 - 9 © 2012 Pearson Education, Inc.. All rights reserved. infinity  lim 1/x as x  infinity ?

© 2012 Pearson Education, Inc.. All rights reserved. Two Tools with Limits

© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 5 Suppose and find Solution:

© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 8 Solution: Here, the highest power of x is x 2, which is used to divide each term in the numerator and denominator.

© 2012 Pearson Education, Inc.. All rights reserved. Your Turn

© 2012 Pearson Education, Inc.. All rights reserved. Section 3.2 Continuity

© 2012 Pearson Education, Inc.. All rights reserved. Figure 14

© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved. Figure

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© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Find all values x = a where the function is discontinuous. Solution: This root function is discontinuous wherever the radicand is negative. There is a discontinuity when 5x + 3 < 0

© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Find all values of x where the piecewise function is discontinuous. Solution: Since each piece of this function is a polynomial, the only x-values where f might be discontinuous here are 0 and 3. We investigate at x = 0 first. From the left, where x-values are less than 0, From the right, where x-values are greater than 0 Continued

© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Continued Because the limit does not exist, so f is discontinuous at x = 0 regardless of the value of f(0). Now let us investigate at x = 3. Thus, f is continuous at x = 3.

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© 2012 Pearson Education, Inc.. All rights reserved. Section 3.3 Rates of Change

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© 2012 Pearson Education, Inc.. All rights reserved. Section 3.4 Definition of the Derivative

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© 2012 Pearson Education, Inc.. All rights reserved.

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© 2012 Pearson Education, Inc.. All rights reserved. Figure 38

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© 2012 Pearson Education, Inc.. All rights reserved. Section 3.5 Graphical Differentiation

© 2012 Pearson Education, Inc.. All rights reserved. Figure 45

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© 2012 Pearson Education, Inc.. All rights reserved. Chapter 3 Extended Application: A Model for Drugs Administered Intravenously

© 2012 Pearson Education, Inc.. All rights reserved. Figure 55

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© 2012 Pearson Education, Inc.. All rights reserved. Figure 57