Curve Sketching. The process of curve sketching can be performed in the following steps:

Transcription

1 Curve Sketching So ar you have learned how to ind st and nd derivatives o unctions and use these derivatives to determine where a unction is:. Increasing/decreasing. Relative extrema 3. Concavity 4. Points o Inlection We will now look at how all o this inormation, in addition to some concepts used in college algebra, can be used to assist in sketching the graph o the unction. The process o curve sketching can be perormed in the ollowing steps:. Determine the domain o the unction (look or any restrictions). Find the x- and y-intercepts or the unction. 3. Locate any asymptotes (vertical, horizontal, oblique) or the unction. 4. Find the st derivative o the unction 5. Locate any critical numbers or the st derivative 6. Determine the intervals where the unction is increasing or decreasing 7. Locate any relative extrema 8. Find the nd derivative 9. Locate any critical numbers or the nd derivative 0. Determine the intervals where the unction is concave up or concave down. Locate any points o inlection. Plot all points (intercepts, relative extrema, inlection points) and asymptotes on the graph. Connect the points with a smooth curve Beore going over an example o using these curve sketching steps, lets irst review a ew o the college algebra steps that you may have orgotten. When determining the domain o unctions there are several common restrictions that you will want to look or. These include:. Division by zero Example: x Restriction: x 0. Taking the square (or other even) root o a negative number Example: n x ; where n is an even number Restriction: x 0 3. Taking the logarithm o zero or a negative number Gerald Manahan SLAC, San Antonio College, 008

2 Example: log x ; b > b 0 but Restriction: x > 0 Example : Find the domain or the unction Solution: x + In this unction you have a raction so you must make sure the denominator is not zero (division by zero). + 0 x The domain o (x) is all real numbers except or /. D ( ) x x x When asked to ind the x- and y-intercepts you will be substituting zero or either x or y. First, lets look at inding the y-intercept since it normally requires less work. The y-intercept is ound by substituting the variable x with zero and then simpliying the expression. ( 0) 4 3x x ( ) The y-intercept or this unction would be at (0, 4) Finding the x-intercept would be done by now letting y o (x) be equal to zero and then solve the expression or x. This will oten require you to use techniques such as actoring or synthetic division. Lets use the same unction as above. Gerald Manahan SLAC, San Antonio College, 008

3 This time we will let (x) 0 4 3x x + 4 3x 0 x + Now in order or this raction to be equal to zero, the numerator (4 3x) must be equal to zero. So you can now set just the numerator equal to zero and solve or x. 4 3x 0 3x 4 x 4 3 The x-intercept or this unction would be at 4,0 3. Example : Find the x- and y-intercepts o the unction x 9 x +. Solution: Find the y-intercept by letting x 0. ( 0) x 9 x + ( ) The y-intercept is (0, 9). Gerald Manahan SLAC, San Antonio College, 008 3

4 Example (Continued): Find the x-intercept(s) by letting (x) 0. 0 x 9 x + x 9 x + Now let the numerator equal zero. x 9 0 x 9 x 9 x 3 x ± 3 The x-intercepts are (-3, 0) and (3, 0). The last concept to review is inding the asymptotes o a unction. A rational unction can have vertical, horizontal, or oblique asymptotes. An important thing to remember is that the unction cannot have both a horizontal and an oblique asymptote. The rational unction will have a horizontal asymptote, an oblique asymptote, or neither. Lets begin with vertical asymptotes. Vertical asymptotes or rational unctions occur whenever the denominator is equal to zero. So to ind the vertical asymptote you will set just the denominator equal to zero and then solve or x. Example 3: Find the vertical asymptote o the unction Solution: Set the denominator equal to zero and solve or x. x 4 0 x 4 The vertical asymptote is x 4. 3x +. x 4 Gerald Manahan SLAC, San Antonio College, 008 4

5 Horizontal asymptotes o a rational unction can be determined by either using limits as x approaches positive and negative ininity or by comparing the highest degree powers o the numerator and denominator. The degree powers o the rational unction can be used to ind the horizontal asymptote by ollowing these rules: Let the rational unction be represented by ax bx I n < m, then the horizontal asymptote will be equal to the x-axis. y 0 I n m, then the horizontal asymptote will be equal to the ratio o the coeicients. a y b I n > m, then there is no horizontal asymptote and you must check or an oblique asymptote. n m Example 4: Find the horizontal asymptote o the unction 3 6x x. 4 x + Solution: Locate the terms with the highest degree power in both the numerator and denominator. 3 6x x x 4 + Compare the degree powers and apply the rules or horizontal asymptotes. The degree power o the numerator (n) 3 The degree power o the denominator (m) 4 n < m, thereore the horizontal asymptote is at y 0. The last asymptote we will discuss is the oblique asymptote. These asymptotes occur or rational unctions whenever the degree power o the numerator is one degree larger than the degree power o the denominator. We will use long division to ind the equation o the oblique asymptote. Gerald Manahan SLAC, San Antonio College, 008 5

6 Example 5: Find the oblique asymptote o the rational unction x 9 x +. Solution: The degree power o the numerator () is one more than the degree power o the denominator (), so we will have an oblique asymptote. x x+ x + 0x 9 ( x x) + x 9 ( x ) 8 The oblique asymptote is y x. Once you know where the unction is increasing or decreasing and its concavity, you can begin to sketch the graph by looking at the our dierent possible graphing combinations. Ok, now lets look at an example, which incorporates all o the steps or curve sketching. Gerald Manahan SLAC, San Antonio College, 008 6

7 Example 6: Graph the unction Solution: Step : Find the domain o (x). 4x using the steps or curve sketching. + Since are unction is a rational unction, the denominator must not be equal to x The domain o (x) is equal to all real numbers except or /. D ( ) x x x Step : Find the x- and y-intercepts. Set x 0 to ind y-intercept. ( 0) 4x ( ) ( ) The y-intercept is at the origin (0, 0). Gerald Manahan SLAC, San Antonio College, 008 7

8 Example 6 (Continued): Step : Find the x- and y-intercepts. Set (x) 0 to ind the x-intercept(s). 4x + 4x 0 + x 0 4x 0 x The x-intercept is also at the origin (0, 0). Step 3: Locate any asymptotes. Set denominator equal to zero to ind vertical asymptotes. + 0 x The vertical asymptote is x. Compare degree powers o the numerator and denominator to ind horizontal asymptote, i any. Degree power o numerator: Degree power o denominator: Degree powers are equal so the horizontal asymptote is equal to the ratio o the coeicients. 4 y y The horizontal asymptote is at y -. Since there is a horizontal asymptote you do not need to look or an oblique asymptote. Gerald Manahan SLAC, San Antonio College, 008 8

9 Example 6 (Continued): Step 4: Find the st derivative. 4x + ( ) x( ) ( ) x( ( + ) + D 4x 4x D + ( + )( 4) ( 4x)( ) ( + ) 4 8x+ 8x ( + ) 4 ( + ) ) Step 5: Locate the critical numbers o the st derivative. Since the numerator o the derivative is a constant the derivative will never be equal to zero. So the only critical numbers or the st derivative will occur when the derivative is undeined (denominator is zero) x The denominator will only be zero i x is /. However, this value is not included in the domain o the unction and is also where our vertical asymptote is located. Thereore, we have no critical numbers or the st derivative. Gerald Manahan SLAC, San Antonio College, 008 9

10 Example 6 (Continued): Step 6: Determine the intervals where the unction is increasing/decreasing. For this unction we have only two intervals to look at, and, Since the numerator o the derivative is the constant (-4) the numerator will always be negative. The denominator is squared so it will always be positive. Thereore, the derivative will always be negative, which means the unction is always decreasing. Intervals where (x) is increasing: Intervals where (x) is decreasing: never, and, Step 7: Locate any relative extrema Since the unction has no critical numbers and is always decreasing, there are no relative extrema. Step 8: Find the nd derivative. Since the numerator o the st derivative is a constant it would be easier to ind the nd derivative by irst rewriting the st derivative in the orm o a power unction 4 ( + ) ( ) 4 ( ) x + x Gerald Manahan SLAC, San Antonio College, 008 0

11 Example 6 (Continued): Step 8: Find the nd derivative. Now use the generalized power rule to ind the nd derivative. ( ) 4 ( ) x + x ( ) 4( )( ) ( ) 3 ( ) ( + ) 3 x + x D x Step 9: Find the critical numbers or the nd derivative. The numerator o the nd derivative is a constant number so the derivative will never be equal to zero, so the only possible critical numbers will come rom the denominator x The only number that will make the denominator zero is /. However, this cannot be a critical number since / is excluded rom the domain o (x). Gerald Manahan SLAC, San Antonio College, 008

12 Example 6 (Continued): Step 0: Determine the intervals where the unction is concave up or concave down. The intervals that we must test are, and,. The unction is concave down on the interval concave up on the interval,, and Step : Locate any points o inlection. The only change in concavity occurs at x -/ but this cannot be a point o inlection since it is not included in the domain o (x). Step : Sketch the graph Now you will plot all o the points and asymptotes that you have ound in the previous steps and then connect the points with a smooth curve. From steps 6 & 0 we know where the unction is increasing/decreasing and its concavity. For the interval, the unction is decreasing and concave down so the graph in this interval will look like For the interval, the unction is decreasing and concave up so the graph in this interval will look like Gerald Manahan SLAC, San Antonio College, 008

13 Example 6 (Continued): Gerald Manahan SLAC, San Antonio College, 008 3