Section 11.4 Tangent Lines and Derivatives

Transcription

1 Section.4 Tangent Lines and Derivatives Te Tangent Problem EXAMPLE: Grap te parabola y x 2 and te tangent line at te point P(,). Solution: We ave: DEFINITION: Te tangent line to te curve y f(x) at te point P(a,f(a)) is te line troug P wit slope m () provided tat tis limit exists. Tere is anoter (equivalent) expression for te slope of te tangent line: m f(a+) f(a) (2)

2 EXAMPLE: Consider te grap of y x 2. (a) Find te slope of te tangent line to te grap at te point P(,). Solution : We ave or m f(a+) f(a) m f(a+) f(a) (a+) 2 a 2 (a+) 2 a 2 a 2 +2a+ 2 a 2 2a+ 2 (2a+) (2a+) 2a+0 2a (a+ a)(a++a) (2a+) (2a+) 2a+0 2a So te slope of te tangent line at te point P(,) is m 2 2. Solution 2: We ave m and te same result follows. (b) Find te equation of te tangent line. x 2 a 2 ()(x+a) (x+a) a+a 2a Solution: Te equation of te tangent line can be found wit te point-slope form of te equation of a line: y y 0 m(x x 0 ) y 2(x ) y 2x 2 y 2x 2+ 2x 2

3 EXAMPLE: Find an equation of te tangent line to te parabola y x 2 8x+9 at te point (3, 6). Solution: We ave m (x 2 8x+9) (a 2 8a+9) x 2 8x+9 a 2 +8a 9 x a x 2 a 2 8x+8a ()(x+a) 8() ()[(x+a) 8] [(x+a) 8] 2a 8 or m f(a+) f(a) [(a+) 2 8(a+)+9] [a 2 8a+9] a 2 +2a+ 2 8a 8+9 a 2 +8a 9 2a+ 2 8 (2a+ 8) (2a+ 8) 2a+0 8 2a 8 So te slope of te tangent line at te point (3, 6) is m Te equation of te tangent line at tis point can be found wit te point-slope form of te equation of a line: y y 0 m(x x 0 ) y ( 6) 2(x 3) y +6 2x+6 y 2x 3

4 Secant lines and tangent lines (or, more precisely, teir slopes) are te geometric analogues of te average and instantaneous rates of cange studied in te previous section, as summarized in te following cart: We ave seen tat limits of te form lim Derivatives or lim f(a+) f(a) arise in finding te slope of a tangent line or te velocity of an object. Moreover, te same type of limit arises wenever we calculate a rate of cange in any of te sciences or engineering, suc as a rate of reaction in cemistry or a marginal cost in economics. Since tis type of limit occurs so widely, it is given a special name and notation. DEFINITION: Te derivative of a function f at a number a, denoted by f (a), is if tis limit exists. REMARK: Equivalently, f (a) f(a+) f(a) f (a) Te derivative function f as as its domain all te points at wic te specified limit exists, and te value of te derivative function at te number x is te number f (x). If y f(x) is a function, ten its derivative is denoted eiter by f or by y. If x is a number in te domain of y f(x) suc tat y f (x) is defined, ten te function f is said to be differentiable at x. Te process tat produces te function f from te function f is called differentiation. Te derivative function may be interpreted in many ways, two of wic were already discussed:. Te derivative function f gives te instantaneous rate of cange of y f(x) wit respect to x. Tis instantaneous rate of cange can be interpreted as marginal cost, marginal revenue, or marginal profit (if te original function represents cost, revenue, or profit, respectively) or as velocity (if te original function represents displacement along a line). From now on, we will use rate of cange to mean instantaneous rate of cange. 2. Te derivative function f gives te slope of te grap of f at any point. If te derivative is evaluated at x a, ten f is te slope of te tangent line to te curve at te point (a,f(a)). 4

5 EXAMPLE: Use te grap of te function f(x) in te Figure below to answer te given questions. y Slope of tangent line is positive Slope of tangent line is negative x Slope of tangent line is 0 (a) Is f (3) positive or negative? Solution: We know tat f (3) is te slope of te tangent line to te grap at te point were x 3. Te Figure above sows tat tis tangent line slants downward from left to rigt, meaning tat its slope is negative. Hence, f (3) < 0. (b) Wic is larger, f () or f ()? Solution: Te Figure above sows tat te tangent line to te grap at te point were x slants upward from left to rigt, meaning tat its slope, f (), is a positive number. Te tangent line at te point were x is orizontal, so tat it as slope 0. (Tat is, f () 0). Terefore, f () > f (). (c) For wat values of x is f (x) positive? Solution: On te grap, find te points were te tangent line as positive slope (slants upward from left to rigt). At eac suc point, f (x) > 0. Te Figure above sows tat tis occurs wen 0 < x < 2 and wen < x < 7. EXAMPLE: If f(x) 3x, find f (x). Solution : We ave f (a) f(a+) f(a) Solution 2: We ave f (a) ( ) 3(a+) (3a ) (3a+3 ) (3a ) 3a+3 3a+ (3x ) (3a ) 3x 3a x 3a 3() 3 3

6 EXAMPLE: If f(x) x, find f (x). Solution : We ave f (a) x a ( x) 2 ( a) 2 ()( x+ a) Solution : We ave a+ a 2 a f (a) Solution 2: We ave f f(a+) f(a) (a) a+) 2 ( a) 2 ( x a x a ( x a)( x+ a) ( a++ a) a++ a Solution 2 : We ave f f(a+) f(a) (a) a+ a ( x a)( x+ a) ()( x+ a) ()( x+ a) x+ a a+ a ( a++ a) a+0+ a a+ a ( a+) 2 ( a) 2 a++ a x a ( x) 2 ( a) 2 x+ a a+ a 2 a ( a+ a)( a++ a) ( a++ a) a+ a 2 a a+ a a+0+ a ( a++ a) a+ a a+ a a+ a ( a+ a)( a++ a) a+ a 2 a EXAMPLE: If f(x) 2x 3 7x, find f (x). Solution: We ave f (a) 2x 3 2a 3 7x+7a (2x 3 7x) (2a 3 7a) 2(x 3 a 3 ) 7() 2x 3 7x 2a 3 +7a ( ) 2(x 2 +xa+a 2 ) 7 2()(x 2 +xa+a 2 ) 7() () ( ) 2(x 2 +xa+a 2 ) 7 2(a 2 +a a+a 2 ) 7 2(3a 2 ) 7 6a 2 7 6

7 EXAMPLE: If f(x) 3 x, find f (x). Solution : We ave f (a) 3 x 3 ( 3 xa a x 3 a xa () ) xa 3 x xa 3 a xa() Solution 2: We ave 3a 3x xa() 3(a x) xa() 3() xa() 3 xa 3 a 2 f (a) f(a+) f(a) 3 a+ 3 a ( 3 (a+)a a+ 3 ) a (a+)a (a+)a 3 a+ (a+)a 3 a (a+)a 3a 3(a+) (a+)a 3a 3a 3 (a+)a 3 (a+)a 3 (a+)a 3 (a+0)a 3 a a 3 a 2 Existence of te Derivative Te definition of te derivative includes te prase if tis limit exists. If te limit used to define f (x) does not exist, ten, of course, te derivative does not exist at tat x. For example, a derivative cannot exist at a point were te function itself is not defined. If tere is no function value for a particular value of x, tere can be no tangent line for tat value. Derivatives also do not exist at corners or sarp points on a grap. Since a vertical line as an undefined slope, te derivative cannot exist at any point were te tangent line is vertical. Tis figure summarizes various ways tat a derivative can fail to exist. f(x) Function not defined Vertical tangent limf(x) x x 3 does not exist. No tangent line possible 0 x x 2 x 3 x 4 x x 7

8 EXAMPLE: Sow tat f(x) x is not differentiable at x 0. Solution: Note tat { x if x < 0 x x if x 0 Terefore on te one and we ave On te oter and, Since f(0+) f(0) lim f(0+) f(0) lim f(0+) f(0) lim + + lim + f(0+) f(0) it follows tat f(0+) f(0) lim does not exist Terefore f (0) does not exist, so f(x) x is not differentiable at x 0. 8

9 Price Elasticity of Demand Any retailer wo sells a product or a service is concerned wit ow a cange in price affects demand for te article. Te sensitivity of demand to price canges varies wit different items. For smaller items, suc as soft drinks, food staples, and ligtbulbs, small percentage canges in price will not affect te demand for te item muc. However, sometimes a small percentage cange in price on big-ticket items, suc as cars, omes, and furniture, can ave significant effects on demand. One way to measure te sensitivity of canges in price to demand is by te ratio of percent cange in demand to percent cange in price. If q represents te quantity demanded and p te price of te item, tis ratio can be written as q/q p/p were q represents te cange in q and p represents te cange in p. Te ratio is usually negative, because q and p are positive, wile q and p typically ave opposite signs. (An increase in price causes a decrease in demand.) If te absolute value of tis quantity is large, it sows tat a small increase in price can cause a relatively large decrease in demand. Applying some algebra, we can rewrite tis ratio as or q/q p/p ( q/q) p q ( p/p) p q ( q) p ( p) q q p p q p q q p q/q p/p q q p p q q p p q p q p p q q p p q q p Suppose q f(p). (Note tat tis is te inverse of te way our demand functions ave been expressed so far; previously, we ad p D(q).) Ten q f(p+ p) f(p). It follows tat As p 0, tis quotient becomes and Te quantity q p f(p+ p) f(p) p q lim p 0 p f(p+ p) f(p) p 0 p dq dp p lim p 0 q q p p q lim q p 0 p p q dq dp E p q dq dp is positive because dq/dp is negative. E is called elasticity of demand and measures te instantaneous responsiveness of demand to price. For example, E may be.6 for pysician services (expenses tat ave considerable price increases eac year, but still ave a ig demand), 9

10 but may be 2.3 for restaurant meals (igly enjoyable, but ig-cost items and not necessities). Tese numbers indicate tat te demand for pysician services is muc less responsive to price canges tan te demand for restaurant meals. Anoter factor tat impacts elasticity of demand is te availability of substitute products. For example, if one airline increases prices on a particular route but oter airlines serving te route do not, ten rater tan paying te iger prices, consumers could fly wit a different airline (te substitute product). If E <, te relative cange in demand is less tan te relative cange in price, and te demand is called inelastic. If E >, te relative cange in demand is greater tan te relative cange in price, and te demand is called elastic. Wen E, te percentage canges in price and demand are relatively equal, and te demand is said to ave unit elasticity. Sometimes elasticity is counterintuitive. Te addiction to illicit drugs is an excellent example. Te quantity of te drug demanded by addicts, if anyting, increases, no matter wat te cost. Tus, illegal drugs are an inelastic commodity. EXAMPLE: Suppose tat te demand for flat screen televisions is expressed by te equation q.02p+20.4 were q is te annual demand (in millions of televisions) and p is te price of te product (in dollars). (a) Calculate and interpret te elasticity of demand wen p $200 and wen p $00. Solution: Since q.02p+20.4, we ave dq/dp.02, so tat E p q dq dp p.02p+20.4 (.02).02p.02p Let p 200 to get.02(200) E.02(200) Since.324 <, te demand was inelastic, and a percentage cange in price resulted in a smaller percentage cange in demand. For example, a 0% increase in price will cause a 3.24% decrease in demand. If p 00, ten.02(00) E.02(00) Since.7 >, te price is elastic. At tis point, a percentage increase in price resulted in a greater percentage decrease in demand. A 0% increase in price resulted in a.7% decrease in demand. 0

11 (b) Determine te price at wic demand ad unit elasticity (E ). Wat is te significance of tis price? Solution : Demand ad unit elasticity at te price p tat made E, so we must solve te equation E.02p.02p p.02p p 20.4 p Demand ad unit elasticity at a price of $409 per flat screen television. Unit elasticity indicates tat te canges in price and demand are about te same. Solution 2: Recall tat We ave Revenue (Price per item) (Number of items) Revenue p q p(.02p+20.4).02p p Te quadratic function R.02p p attains its maximum value if wic gives us te same result follows. p b 2a (.02) REMARK: Revenue is maximized wen price is set so tat elasticity of demand is exactly one.

12 Appendix EXAMPLE: If f(x) 3 2x 7x 2, find f (x). Solution : We ave f (a) f(a+) f(a) Solution 2: We ave f (a) ) ) (3 2(a+) 7(a+) 2 (3 2a 7a 2 3 2a 2 7(a 2 +2a+ 2 ) 3+2a+7a 2 3 2a 2 7a 2 4a a+7a 2 2 4a 7 2 ( 2 4a 7) ( 2 4a 7) 2 4a a (3 2x 7x 2 ) (3 2a 7a 2 ) 3 2x 7x 2 3+2a+7a 2 2x 7x 2 +2a+7a 2 2x+2a 7x 2 +7a 2 2() 7(x 2 a 2 ) 2() 7()(x+a) ( ) () 2 7(x+a) ( ) 2 7(x+a) 2 7(a+a) 2 7(2a) 2 4a 2

13 EXAMPLE: If f(x) x+9, find f (x). Solution : We ave In sort, f (a) ( x+9 a+9 x+9 a+9)( x+9+ a+9) ()( x+9+ a+9) ( x+9) 2 ( a+9) 2 ()( x+9+ a+9) (x+9) (a+9) ()( x+9+ a+9) x+9 a 9 ()( x+9+ a+9) x a ()( x+9+ a+9) () ()( x+9+ a+9) x+9+ a+9 a+9+ a+9 2 a+9 f (a) x+9 a+9 ( x+9 a+9)( x+9+ a+9) ()( x+9+ a+9) ( x+9) 2 ( a+9) 2 ()( x+9+ a+9) (x+9) (a+9) ()( x+9+ a+9) () ()( x+9+ a+9) x+9+ a+9 2 a+9 3

14 Solution 2: We ave f (a) f(a+) f(a) In sort, f (a) ( ( (a+)+9 a+9 (a+)+9 a+9)( (a+)+9+ a+9) ( (a+)+9+ a+9) ( (a+)+9) 2 ( a+9) 2 ( (a+)+9+ a+9) ((a+)+9) (a+9) ( (a+)+9+ a+9) a++9 a 9 ( (a+)+9+ a+9) ( (a+)+9+ a+9) (a+)+9+ a+9 (a+0)+9+ a+9 a+9+ a+9 2 a+9 (a+)+9 a+9 (a+)+9 a+9)( (a+)+9+ a+9) ( (a+)+9+ a+9) ( (a+)+9) 2 ( a+9) 2 ( (a+)+9+ a+9) ( (a+)+9+ a+9) ((a+)+9) (a+9) ( (a+)+9+ a+9) (a+)+9+ a+9 2 a+9 4

15 EXAMPLE: If f(x) 2 3x, find f (x). Solution: We ave In sort, f (a) ( 2 3x 2 3a 2 3x 2 3a)( 2 3x+ 2 3a) ()( 2 3x+ 2 3a) ( 2 3x) 2 ( 2 3a) 2 ()( 2 3x+ 2 3a) (2 3x) (2 3a) ()( 2 3x+ 2 3a) 2 3x 2+3a ()( 2 3x+ 2 3a) 3x+3a ()( 2 3x+ 2 3a) 3() ()( 2 3x+ 2 3a) 3 2 3x+ 2 3a 3 2 3a+ 2 3a a f (a) 2 3x 2 3a ( 2 3x 2 3a)( 2 3x+ 2 3a) ()( 2 3x+ 2 3a) ( 2 3x) 2 ( 2 3a) 2 ()( 2 3x+ 2 3a) (2 3x) (2 3a) ()( 2 3x+ 2 3a) 3() ()( 2 3x+ 2 3a) x+ 2 3a 2 2 3a

16 EXAMPLE: If f(x) x+3 4 x, find f (x). Solution : We ave f (a) x+3 4 x a+3 4 a x+3 a+3 (4 x)(4 a) 4 x 4 a (4 x)(4 a) () (4 x)(4 a) (x+3)(4 a) (a+3)(4 x) ()(4 x)(4 a) 4x xa+2 3a 4a+ax 2+3x ()(4 x)(4 a) Solution 2: We ave 7() ()(4 x)(4 a) f f(a+) f(a) (a) ) ( a++3 4 a a+3 4 a (4 a )(4 a) ( x+3 4 x a+3 4 a ) (4 x)(4 a) () (4 x)(4 a) (4x xa+2 3a) (4a ax+2 3x) ()(4 x)(4 a) 7x 7a ()(4 x)(4 a) 7 (4 x)(4 a) 7 (4 a)(4 a) 7 (4 a) 2 (a+)+3 4 (a+) a+3 4 a (4 a )(4 a) a++3 a+3 (4 a )(4 a) 4 a 4 a (4 a )(4 a) (4 a )(4 a) (a++3)(4 a) (a+3)(4 a ) (4 a )(4 a) a++3 4 a a+3 4 a (4a a 2 +4 a+2 3a) (4a a 2 a+2 3a 3) (4 a )(4 a) 7 (4 a )(4 a) 7 (4 a )(4 a) 7 (4 a 0)(4 a) 7 (4 a) 2 6