MATH 12002 - CALCULUS I 3.3: Graphing Example Professor Donald L. - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §3.3: Graphing Example

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Example

Example

Let f (x) = 3

2x2/3 − 3 5x5/3.

Determine intervals where f is increasing, intervals where f is decreasing, the location of all local maxima and minima, intervals where f is concave up, intervals where f is concave down, and the location of all inflection points.

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Example

We need to determine the signs of f ′ and f ′′ for f (x) = 3

2x2/3 − 3 5x5/3.

First, f ′(x) = x−1/3 − x2/3 = 1 x1/3 − x2/3 = 1 − (x1/3)(x2/3) x1/3 = 1 − x x1/3 . Hence f ′(x) = 0 when x = 1 and f ′(x) is undefined when x = 0.

D.L. White (Kent State University) 3 / 14

Example

Using f ′(x) = x−1/3 − x2/3, we have f ′′(x) = −1

3x−4/3 − 2 3x−1/3

= −1 3 1 x4/3 + 2 x1/3

• =

−1 3 1 + 2x x4/3

• =

−1 3

• 2(1

2 + x)

x4/3

• =

−2(x + 1

2)

3x4/3 . Hence f ′′(x) = 0 when x = −1

2 and f ′′(x) is undefined when x = 0.

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Example

Notice that f ′(x) = 1−x

x1/3 has a nonlinear factor x1/3 =

3

√x and f ′′(x) =

−2(x+ 1

2 )

3x4/3

has a nonlinear factor x4/3 =

3

√ x4 = ( 3 √x)4. How do we determine the signs of these factors? If a is a real number and n is an odd integer, a and an have the same sign. Applying this to a = x1/3 and n = 3, we see that x1/3 and (x1/3)3 = x have the same sign. More generally, we have If a is a real number and n is an odd integer, then a, an, and a1/n =

n

√a all have the same sign. Now x4/3 = (x1/3)4, so x4/3 is always positive (or 0).

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Example

f (x) = 3

2x2/3 − 3 5x5/3, f ′(x) = 1−x x1/3 , f ′′(x) = −2(x+ 1

2 )

3x4/3

− 1

2

1 1 − x x1/3 f ′(x) −2 x + 1

2

3x4/3 f ′′(x)

Inc-Dec Concave Shape

+ + + − − − + + X − − + − − − − − − + + + + + + + X + − − − ✲ ✲ D D I D ✛ ✛ U D D D ✍ ☞ ✎ ☞

INF MIN MAX D.L. White (Kent State University) 6 / 14

Example

f (x) = 3

2x2/3 − 3 5x5/3, f ′(x) = 1−x x1/3 , f ′′(x) = −2(x+ 1

2 )

3x4/3

− 1

2

1

Inc-Dec Concave Shape

D D I D

MIN MAX

U D D D

INF

✡ ✟ ☛ ✟

f is increasing on (0, 1), f is decreasing on (−∞, 0) ∪ (1, ∞); f has a local minimum at x = 0, f has a local maximum at x = 1. f is concave up on (−∞, −1

2), f is concave down on (−1 2, 0) ∪ (0, ∞);

f has an inflection point at x = −1

2.

D.L. White (Kent State University) 7 / 14

Example

In order to sketch the graph of f , we will need to plot the points whose x coordinates are in the sign chart. We need to evaluate f (x) = 3

2x2/3 − 3 5x5/3 at these points:

f (−1

2)

=

3 2(−1 2)2/3 − 3 5(−1 2)5/3

= (−1

2)2/3 3 2 − 3 5(−1 2)

• =
• 1

(−2)2/3

3

2 + 3 10

• =
• 1

41/3

15

10 + 3 10

• =

1

3

√ 4 · 18 10 = 9 5 3 √ 4 ≈ 1.13

f (0) =

3 2(0)2/3 − 3 5(0)5/3 = 0

f (1) =

3 2(1)2/3 − 3 5(1)5/3 = 3 2 − 3 5 = 15 10 − 6 10 = 9 10 = 0.9.

Hence the points (−1

2, 9 5 3 √ 4) ≈ (−0.5, 1.13), (0, 0), and (1, 9 10) = (1, 0.9)

are on the graph.

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Example

It is also useful to have the x-intercepts; that is, x values where f (x) = 0. We can factor f (x) as follows: f (x) =

3 2x2/3 − 3 5x5/3

= 3x2/3 1

2 − 1 5x3/3

= 3x2/3 1

2 − 1 5x

• .

Therefore, f (x) = 0 when x2/3 = 0, and so x = 0 (also the y-intercept),

• r when 1

2 − 1 5x = 0, and so x = 5 2 = 2.5.

Hence the x-intercepts are x = 0 and x = 2.5.

D.L. White (Kent State University) 9 / 14

Example

− 1

2

1

Inc-Dec Concave Shape

D D I D

MIN MAX

U D D D

INF

✡ ✟ ☛ ✟ ✲ ✛ ✻ ❄

1 2 −1 1 2 3

q

(− 1

2 , ≈ 1.13), INF

q

(0, 0), MIN

q

(1, 0.9), MAX

q

(2.5, 0), INT D.L. White (Kent State University) 10 / 14

Example

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Example

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Example

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Example

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