Using the MVT: Increasing and Decreasing Functions

Transcription

1 Using the MVT: Increasing and Decreasing Functions F irst let s be clear on what and decreasing functions are DEFINITION 317 Assume f is defined on an interval I f is on I if whenever a and b are in I and a < b, then f (a) < f (b) YOU TRY IT 38 How would you modify the definition to describe f is decreasing on I? As noted, the key value of the MVT is in roving other results THEOREM 318 (Increasing/Decreasing Test) 1 If (x) > for all x in an interval I, then f is on I 2 If (x) < for all x in an interval I, then f is decreasing on I Answer to you try it 38 f is decreasing on I if whenever a and b are in I and a < b, then f (a) > f (b) Proof We ll rove (2) So assume (x) < for all x in I Let a and b be any two oints in I with a < b To show that f is decreasing, we need to show that f (b) < f (a) a But f is differentiable on I so it is continuous on [a, b] and differentiable on (a, b) so the MVT alies to f on [a, b] So there s a oint c in (a, b) so that b or f (b) f (a) = (c) b a f (b) f (a) = (c)(b a) =( )(+) < But f (b) f (a) < ) f (b) < f (a) which is what we wanted to show YOU TRY IT 39 Prove case (1) of the Increasing/Decreasing Test where (x) > EXAMPLE 319 Let f (x) =x 4 have relative extrema? 6x Where is f? Decreasing? Where does it Solution Use the Increasing/Decreasing Test Find the derivative and the critical oints ( (x) = or DNE) (x) =4x 3 12x = 4x(x 2 3)=4x(x 3)(x + 3)= at x = ± 3,

2 math 13, day 2 the first derivative test 12 Now record this information on a number line for easy reference (, 3, ) (, 3, ) (3 1/2 ) 3 1/2 Now determine the sign of (x) between and beyond the critical oints Here we use the IVT to know that the only laces that the derivative can change sign are at the critical oints because the derivative is continuous Just lug in values in the aroriate intervals ( 2) = 8, ( 1) =+8, (1) = 8, and (2) =+8 (, 3, ) (, 3, ) decreasing decreasing (3 1/2 ) 3 1/2 3, ) and ( 3, ) and it is decreas- Using interval notation: f is on ( ing on (, 3) and (, 3) EXAMPLE 311 (This examle is also used in the next section) Let f (x) =xe 2x Where is f? Decreasing? Use the Increasing/Decreasing Test Find the derivative and the criti- Solution cal oints (x) =e 2x + 2xe 2x = e 2x [1 + 2x] = at x = 1/2 Set u the number line and determine the sign of (x) on either side of the critical oint ( 1) = e 2 < and () =1 decreasing 1/2 Using interval notation: f is on ( (, 1/2) 1/2, ) and it is decreasing on EXAMPLE 3111 [This examle is also used in the next section] Let f (x) =x 2 e x Where is f? Decreasing? Use the Increasing/Decreasing Test Find the derivative and the criti- Solution cal oints (x) =2xe x + x 2 e x = e x [2x + x 2 ]=e x (x)(2 + x) = at x =, 2 Set u the number line and determine the sign of (x) on either side of the critical oints ( 3) =e 3 ( 3)( 1) > ( 1) =e 1 ( 1)(+1) < (1) =e 1 (1)(3) > f decreasing Using interval notation: f is on (, 2) [ (, ) and it is decreasing on ( 2, )

3 The First Derivative Test B ecause we know that (x) > ) f is (x) < ) f is decreasing, we can use this to classify a critical oint as either local max or local min or neither as we did in the last examle < < = loc min DNE loc min > > > > = loc max DNE loc max < < < > = neither DNE neither > < Figure 295: If the sloe (x) changes sign on either side of a critical oint of f, what can you say about that oint? Using the intuition from the Increasing/Decreasing Test, we obtain: THEOREM 2911 (The First Derivative Test) Let c be a critical oint of a continuous function f If changes sign from ositive to negative at c, then f has a local max at c If changes sign from negative to ositive at c, then f has a local min at c If does not changes sign at c, then f has neither a local max nor min at c YOU TRY IT 291 Return to the examle at the end of the last section and classify the critical oints using the First Derivative Test EXAMPLE 2912 Classify the relative extrema of f (x) =x 4 6x Solution We determined earlier the sign of (x) along a number line So now all we have to do is fill in the tye of critical oint dec r min inc r max dec r min inc (3 1/2 ) 3 1/2 EXAMPLE 2913 Classify the relative extrema of h(x) =(x 2 grah lotting only the critical oints 4) 3 Then sketch a quick

4 math 13, day 2 the first derivative test 14 Solution Use the First Derivative Test h (x) =3(x 2 4) 2 (2x) = at x =, ±2 It is easy to determine the sign of the derivative dec neither dec r min inc neither inc h 2 2 h 2 2 We can make a sketch of the grah by lotting the critical oints and using the sloe information on the number line Check that h(2) =, h() = 64, and h( 2) = (, 64) YOU TRY IT 2911 Classify the relative extrema of f (x) = 1 4 (x2 4) 2 answer to you try it 2911 The critical oints are at x = (relative max) and ±2 (both relative mins) EXAMPLE 2914 Classify the critical oints f (x) = x 2 e x that we examined in Examle 3111 Solution We found that f decreasing Consequently, there is a relative max at x = 2 where f (x) =4e There is a relative min at x = where f () = We can lot these two oints and make an educated guess at the shae of the grah using the information in the number line f (x) =x 2 e x 3 2 rel max: ( 2, 541) 1 2 rel min: (, ) EXAMPLE 2915 Suose that a function f (x) is continuous and has the following number line that describes its first derivative Interret this information to find where f is, decreasing, and has relative extrema Then draw a grah of the original function f that satisfies these conditions Solution have DNE 3 3 Use the First Derivative Test to determine what tye of extrema we dec r min inc r max dec r min inc DNE 3 3

5 math 13, day 2 the first derivative test 15 We can use this information to grah one ossible solution for f Note the function is cannot be differentiable at (but should be elsewhere) 3 3 EXAMPLE 2916 Suose that the grah of is given below Translate this information into number line form and then attemt to grah the original function f Solution All we need to do is ay attention to the sign of the derivative dec r min inc r max dec none dec 4 4 Here s one function that satisfies these conditions Notice that x = 4 is not a relative extreme oint More Examles EXAMPLE 2917 Let f (x) =xe 2x Where is f? Decreasing? Where does it have relative extrema? Use the Increasing/Decreasing Test Find the derivative and the criti- Solution cal oints (x) =e 2x + 2xe 2x = e 2x [1 + 2x] = at x = 1/2

6 math 13, day 2 the first derivative test 16 Set u the number line and determine the sign of (x) on either side of the critical oint ( 1) = e 2 < and () =1 decreasing r min 1/2 Using interval notation: f is on ( 1/2, ) and it is decreasing on (, 1/2) From the First Derivative Test, there is a relative min at x = 1/2 EXAMPLE 2918 Let f (x) =x have relative extrema? sin x Where is f? Decreasing? Where does it Use the Increasing/Decreasing Test Find the derivative and the criti- Solution cal oints (x) =1 cos x = at x =, ±2, ±4 Since cos x ale 1 the sign of (x) between the critical oints is always ositive So the f (x) is always and by the First Derivative Test and there are no relative extrema f increase Not Extr increase Not Extr increase Not Extrincrease EXAMPLE 2919 Let f (x) =(x 2 have relative extrema? 4) 2/3 Where is f? Decreasing? Where does it Use the Increasing/Decreasing Test Find the derivative and the criti- Solution cal oints (x) = 2 3 (x2 4) 1/3 2x = 4x 3(x 2 = at x = DNE at x = ±2 4) 1/3 Determine the sign of (x) between and beyond the critical oints ( 3) <, ( 1) >, (1) < and (3) > decr r min incr r max decr r min incr DNE +++ DNE Using the First Derivative Test, f has relative mins at x = ±2 and a relative max at Can you sketch the shae of the grah based on the information? R max: (, 25198) Figure 296: Evaluate f at the critical oints and lot them Then connect these oints using the sloe information on the number line f ( 2) = f (2) = f () =( 4) 2/3 = 16 1/ DNE DNE R min R min