4
1 Example 4 Sketch the graph of the function k(x) = (x 2 - 4) 4/5 . Solution Observe that k is an even function, and its graph is symmetric with respect to the y-axis. I. Intercepts The x-intercepts occur when 0 = x 2 -4 , i.e. when x=-2 and when x=2. The y-intercept occurs at II. Asymptotes The graph of k has no asymptotes. III. First Derivative By the chain rule, 5 5 5 1 5 4 8 2 256 256 4 0 0 k( ) ( ) / / . ) ( ) ( ) ( ) ( ) ( ) / / / / 5 1 5 1 5 1 2 5 1 2 2 x 2 x 5 x 8 4 x 5 x 8 x 2 4 x 5 4 (x k Since k / (x)>0 for –2<x<0 and x>2, the function k is increasing there. Since k / (x)<0 for x<-2 and 0<x<2, the function k is decreasing there. We depict this information on a real number line.

1 Example 4 Sketch the graph of the function k(x) = (x 2 -4) 4/5. Solution Observe that k is an even function, and its graph is symmetric with respect

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Page 1: 1 Example 4 Sketch the graph of the function k(x) = (x 2 -4) 4/5. Solution Observe that k is an even function, and its graph is symmetric with respect

1

Example 4 Sketch the graph of the function k(x) = (x2-4)4/5.

Solution Observe that k is an even function, and its graph is symmetric with respect to the y-axis.

I. Intercepts

The x-intercepts occur when 0 = x2-4 , i.e. when x=-2 and when x=2.

The y-intercept occurs at

II. Asymptotes The graph of k has no asymptotes.

III. First Derivative

By the chain rule,

555154 82256256400k( )() //

.)()()(

)()() ////

5151512512

2x2x5x8

4x5x8

x24x54

(xk

Since k/(x)>0 for –2<x<0 and x>2, the function k is increasing there. Since k/(x)<0 for x<-2 and 0<x<2, the function k is decreasing there. We depict this information on a real number line.

Page 2: 1 Example 4 Sketch the graph of the function k(x) = (x 2 -4) 4/5. Solution Observe that k is an even function, and its graph is symmetric with respect

2

decr -2 i n cr 0 decr 2 i n cr

si gn k ’(x )x

- - - - D N E + + + + 0 - - - - - - - D N E + + + + +

loca lm i n

l oca lm i n

l oca lm ax

Note that k has three critical points: x=0 where the derivative is zero and x=-2, x=2 where the derivative does not exist. By the First Derivative Test x=-2 is a local minimum, x=0 is a local maximum and x=2 is a local minimum.

IV. Vertical Tangents and Cusps

Observe that at x=-2 the left derivative of k is - and the right derivative is +, and k has a vertical cusp there. Similarly, at x=2 the left derivative of k is -and the right derivative is +, and k also has a vertical cusp there.

V. Concavity and Inflection Points

By the quotient rule:

5/65/65/65/6

2

5/65/6

2

5/62

22

5/42

5/42

5/22

5/425/12

)2()2(25

)203)(203(8

)2()2(25

)203(8

)2()2(25

16024

)4(25

16)4(40

)4(5

)4(5

)4(5

)2()4(51

8)4)(8()(

xx

xx

xx

x

xx

x

x

xx

x

x

x

xxxxxk

5151 2x2x5x8

(xk // )()()

Page 3: 1 Example 4 Sketch the graph of the function k(x) = (x 2 -4) 4/5. Solution Observe that k is an even function, and its graph is symmetric with respect

3

VI. Sketch the graph

We summarize our conclusions and sketch the graph of k.

Since the concavity of k changes from up to down at there is an inflection point there. Since the concavity of k changes from down to up at there is an inflection point there.

The denominator of k//(x) is always positive, so k//(x) has the same sign as

its numerator. Hence k//(x) is positive for

and is concave up there. k//(x) is negative for and is concave down there. We sketch this information on a number line.

5/65/6 )2()2(25

)203)(203(8)(

xx

xxxk

320 and 320 xx

320320 x

,320x

,320x

320 3

20

Page 4: 1 Example 4 Sketch the graph of the function k(x) = (x 2 -4) 4/5. Solution Observe that k is an even function, and its graph is symmetric with respect

4

k is an even function x-intercepts: x=-2 and x=2 y-intercept:

increasing: –2<x<0 or x>2 decreasing: x<-2 or 0<x<2

local min: x=-2 and x=2 local max: x=0 vertical cusps: x=-2 and x=2

concave up: concave down:

inflection points:

5 82y

k(x) =(x2-4)4/5

320320 x320 and 320 xx

320 and 320 xx

3

20

3

20

5 82