Curve Sketching

CURVE SKETCHING I. THE FIRST DERIVATIVE TEST: Let's take an arbitrary function like the one whose graph is given below:

As x goes from a to p, the graph rises as x moves to the right towards the interval P, (a, p) and the value of f increases. We say that the function is increasing on the open interval (a, p). As x continues to move towards q the graph falls and the value of f decreases, we would say that the open interval (p,q) is decreasing. Since f(r) ≥ f(x) for all x in (a, b), we say that the function f has its maximum value on (a, b) when x = r and that f(r) is the maximum value of f on interval (a, b). Notice also that f(q) ≤ f(x) for all x in interval (a, b), we say that the function ƒ has its minimum value on interval (a, b) when x = q and that f(q) is the minimum value of f on interval (a, b). Example 1: Find the critical numbers of f if f(x) = 2x3 - 6x2 - 18x + 7 f '(x)= 6x2 - 12x -18 6x2 - 12x -18 = 0 (6x + 6) (x - 3) = 0 6x + 6 = 0 or x = -1 or

x-3=0 x=3

find f '(x) and set it = 0 factor solve for x

So, the critical numbers are x = -1 and x = 3.

Example 2: Find the intervals on which ƒ is increasing if f(x) = 2x3 - 6x2 - 18x + 7 f is increasing if f ' is positive, so we need to find when f '(x) >0 or 6x2 - 12x - 18 > 0. From Example 1 we know that the critical points are x = -1 and x = 3.

We determine the sign of f ' on the intervals between critical points by using test numbers that fall to the left, between, and to the right of our critical points and plug them into f '(x). (a) From the interval (- ∞ , -1) we choose x = -2 as a test number: f ' (-2) = 6(-2)2 - 12(-2) - 18 = 30 > 0

f is increasing on (- ∞ , -1)

(b) From the interval (-1, 3) we choose x = 0 as a test number: f ' (0) = 6(0)2 - 12(0) - 18 = -18 < 0 f is decreasing on (-1, 3) (c) From the interval (3, ∞ ) we choose x = 4 as a test number: f in increasing on (3, ∞ ) f ' (4) = 6(4)2 -12(4) - 18 = 30 > 0 From (a), (b), and (c) we see that f is increasing on the intervals (- ∞ , -1) and (3, ∞ ).

Example 3: Find the relative extreme values of f (maximum and minimum) if f(x) = 2x3 - 6x2 - 18x + 7 Make the following chart from the results of Example 2: +

Sign of f'

-1

+ 3

Notice that f ' changes sign from (+) to (-) at x = -1 ⇒ f has a relative maximum at x = -1 The value of the relative maximum is f (-1) = 17 f ' changes sign from (-) to (+) at x = 3 ⇒ f has a relative minimum at x = 3 The value of the relative minimum is f (3) = -47

Example 4: Sketch the graph of f if f (x) = 2x3 - 6x2 - 18x + 7 • • • • •

ALWAYS start left to right when graphing. Plot the maximum and minimum points: (-1, 17) and (3, -47) f is increasing on (- ∞ , -1); the graph increases until x = -1 f is decreasing on (-1, 3); the graph decreases until x = 3 f is increasing again on (3, ∞ ); the graph increases until x = ∞ :

II. CONCAVITY AND THE SECOND DERIVATIVE TEST: A graph can be either increasing or decreasing, it could concave up, down, both or neither:

The concavity of the curve has nothing to do with whether the function is increasing or decreasing. See above.

2

Inflection Point: A point is said to be an inflection point if the graph changes concavity at that point. The Second Derivative Test: If f '(c) = 0 (c is a critical number) and f ''(c) > 0, then f has a relative minimum at x = c and is concave up f ''(c) < 0, then f has a relative maximum at x = c and is concave down f ''(c) = 0, the c is an inflection point and changes concavity at x = c. Example 5: Use the second derivative test to find the relative extreme values of f if f (x) = 2x3 - 6x2 - 18x + 7 f '(x) = 6x2 - 12x - 18 The critical numbers are x = -1 and x = 3 f ''(x) = 12x -12

(see Example 1)

Now evaluate f ''(x) at the critical numbers: f ''(-1) = 12(-1) -12 = -24 < 0 ⇒ relative maximum at x = -1 f ''(3) = 12(3) -12 = 24 > 0 ⇒ relative minimum at x = 3 (see Example 2)

Example 6: Use the second derivative test to find the relative extreme values and the inflection points of f if f (x) = 2 - 9x + 6x2 - x3 f '(x)= -9 + 12x - 3x2 -9 + 12x - 3x2 = 0 -3(x-3)(x-1) = 0 x=3 or x=1

find f '(x) and set it = 0 factor solve for x

The critical numbers are: x = 1 and x = 3. Sign-of-f'

-

+

-

1

3

f is decreasing on both (- ∞ , 1) and (3, ∞ ) and increasing on (1, 3). f (1) = -2 and f(3) = 2 Plot the points (1, 2) and (3, 2) on the graph. To discuss concavity look at f '': f ''(x) = 12 - 6x = 0 x=2 • We determine the sign of f '' by choosing points that lie to the left, between or to the right of our inflection points and plugging them into f ''(x). • For interval (- ∞, 2) we choose x = 0; f ''(0) = 12-6(0) = 12 > 0. • • • •

•

For interval ( 2, ∞ ) we choose x = 4; f ''(4) = 12-6(4) = -12< 0.

Sign-of-f ''

+

2

• • •

the graph is concave up on (- ∞ , 2) and concave down on (2, ∞ ) x = 2 is an inflection point; f(2) = 2 - 9(2) + 6(2)2 - (2)3 =0: x = 2 is also the x-intercept; f(2) = 0

3

III. ASYMPTOTES: Asymptotes exist only when the function consists of fractional polynomials. We have three types of asymptotes; vertical, horizontal and oblique (slant): Vertical Asymptotes: Whatever number makes the denominator zero is a vertical asymptote: x 3 − 5x 2 + 6 x + 5 i.e., f(x) = , factor x2 – 3x -4 down to (x – 4)(x + 1), 2 x − 3x − 4 x = 4 and x = -1 are vertical asymptotes.

Horizontal Asymptotes Look at the highest power of the numerator (call it t for top) and the highest power of the denominator (call it b for bottom); three cases are considered. Case 1: b > t The line y = 0 is a horizontal asymptote. Case 2.: b = t Divided the coefficient of the xt term by the coefficient of the xb term; the result is the horizontal asymptote. Case 3: b + 1 = t Divide the numerator by the denominator and the quotient is called an oblique (slant) asymptote. Otherwise: No asymptotes exist.

Example 7: Find the relative extreme values, the inflection points of f and sketch the graph of: f (x) =

6− x x2

f '(x) = x − 12 = 0 x3 x - 12 = 0 x = 12 •

find f '(x) and set it = 0 set the numerator only = 0 solve for x The critical numbers are: x = 12 and x = 0 (denominator cannot = 0, so that value is considered a critical number)

4

• f is increasing on both (- ∞ , 0) and (12, ∞ ) and decreasing on (0, 12). • f(0) = undefined and f(12) = -6/144 • the point (12, -6/144) is a relative minimum. • Plot the point (12, -6/144) on the graph. • To discuss concavity look at f '': f ''(x)= − 2 x + 36 = 0 x4 x = 18 and again x = 0 are critical numbers

+

Sign-of-f '' • • • • •

+ 0

18

the graph is concave up on (- ∞ , 0) and (0, 18) and concave down on (18, ∞ ) x = 18 is the only inflection point: Vertical asymptotes: x = 0 (denominator = 0) Horizontal asymptotes: y = 0 (second case) and finally the graph:

Revised: Spring 2005 Created in 1994 by Ziad Diab STUDENT LEARNING ASSISTANCE CENTER (SLAC) Texas State University-San Marcos

5