Differentiation 9I a f ( ) f ( ) 6 6 f ( ) 6 ii f ( ) is concave when f ( ) sin for π So f ( ) is concave on the interval [, π]. i f ( ) is conve when f ( ) 6 6 for So f ( ) is conve for all or on the interval [, ). f ( ) 7 f ( ) f ( ) ii f ( ) is concave when f ( ) 6 6 for So f ( ) is concave for all or on the interval (,]. 4 f ( ) f ( ) 4 9 f ( ) 8 6 ( ) i f ( ) is conve when f ( ) e i f ( ) is conve when f ( ) So f ( ) is not conve anywhere. ii f ( ) is concave when f ( ) So f ( ) is concave for all or on the interval (, ). f ( ) e f ( ) e f ( ) e 6( ) for or So f ( ) is conve for or, i f ( ) is conve when f ( ) e for In So f ( ) is conve on [ln, ). or on (,],. ii f ( ) is concave when f ( ) ii f ( ) is concave when f ( ) e for In So f ( ) is concave on (, ln ]. 6( ) for So f ( ) is concave for all or on the interval c f ( ) sin f ( ) cos f ( ) sin,. i f ( ) is conve when f ( ) sin for π π So f ( ) is conve on the interval [π, π]. f f ( ) ln, f ( ) f ( ) i f ( ) is conve when f ( ) But for all So f ( ) is not conve anywhere. ii f ( ) is concave when f ( ) for all So f ( ) is concave on (, ). Pearson Eucation Lt 7. Copying permitte for purchasing institution only. This material is not copyright free.
a f ( ) arcsin f ( ) ( ) f ( ) ( ) ( ) ( ) On the interval (, ), < f ( ) So f ( ) is concave on the interval (, ). On the interval (, ), > f ( ) So f ( ) is conve on the interval (, ). c f ( ) changes from concave to conve at = When =, y =. point of inflection is (, ). a f ( ) cos sin f ( ) cos sin cos f ( ) (cos sin ) sin ( sin ) sin 4sin sin (sin sin ) At points of inflection f ( ) sin sin (sin )(sin ) sin or π 5π π, or 6 6 Check the sign of f ( ) on either sie of each point: f () ( ) π f ( ) π is an inflection point 6 f(π) ( ) 5π is an inflection point 6 f (π) ( ) π is not an inflection point π y 6 4 5π y 6 4 So the points of inflection are π, 6 4 an 5π, 6 4. Pearson Eucation Lt 7. Copying permitte for purchasing institution only. This material is not copyright free.
f ( ) f ( ) ( ) f ( ) ( ) ( ) At points of inflection f ( ) ( ) ( ) Check the sign of f ( ) on either sie of = : f (.5).47... f (.5) 4 is a point of inflection When, y ( ) So the point of inflection is (, ). c f ( ) 4 f ( ) ( ) ( ) 4 4 f ( ) ( ) ( ) 4 ( ) ( ) At points of inflection f ( ) ( ) ( ) ( ) ( ) ( ) Check the sign of f ( ) on either sie of = : 4 f ( ) 7 4 f () 7 is a point of inflection When, y So the point of inflection is (, ). f ( ) arctan f ( ) f ( ) ( ) At points of inflection f ( ) only when ( ) When, f ( ) When, f ( ) y So the point of inflection is (, ). Pearson Eucation Lt 7. Copying permitte for purchasing institution only. This material is not copyright free.
4 5 a f ( ) ln f ( ) 4ln ( ln ) f ( ) ( ln ) 6 4ln At a point of inflection f ( ) 6 4ln ln e So there is one point of inflection, where e y e ( ) e ( ) e ( ) e At stationary points e when an stationary point at (, ) e e e ( ) When, so is neither a maimum nor a minimum point When, When, (, ) is a stationary point of inflection. At points of inflection e ( ) or From part a it is known that = is a stationary point of inflection. When, When, so is a point of inflection y e (,e ) is a non-stationary point of inflection. 6 a y e e e e ( ) At stationary points e ( ) when an e stationary point at, e e e ( ) e ( ) When, e Therefore, is a minimum. e At points of inflection e ( ) c, y e e When, When, so is a point of inflection non-stationary point of inflection at, e 7 i f ( ) is the graient, so it is negative for A zero for B positive for C zero for D Pearson Eucation Lt 7. Copying permitte for purchasing institution only. This material is not copyright free. 4
7 ii f ( ) etermines whether the curve is conve, is concave or has a point of inflection. Hence f ( ) is positive for A positive for B negative for C zero for D 8 f ( ) tan 9 a f ( ) sec sin f ( ) sec tan cos At points of inflection f ( ) sin only when sin, cos which has only one solution, =, in π π the interval When, f ( ) When, f ( ) When, f ( ) there is one point of inflection at (, ). 5 y ( ) 5 ( ) ( ) 4 () (8 ) 4 5 ( ) (8 ) 8( ) 4 ( ) (8 ) 8( ) ( ) (9 ) a ( 5) for all, so even though at = 5, the sign of oes not change on either sie of = 5 an hence it is not a point of inflection. 4( 5) when 5 an Stationary point is at (5, ). When 5, When 5, (5, ) is a minimum point. y ln 5 ln ln ln ln C is conve when ln In e At points of inflection () (9) or 9 5 y 9 9 87 y 5 Points of inflection are, an, 9 87 Pearson Eucation Lt 7. Copying permitte for purchasing institution only. This material is not copyright free. 5
Challenge A general cuic can e written as f ( ) a c a c f ( ) 6a f ( ) f ( ) when Let ; then a f 6a a a 6a 6a f 6a a a 6a 6a The sign of f ( ) changes either sie of, so this is the single point of a inflection. a 4 y a c e 4a c a 6 c a 6 c As this is a quaratic equation, there are at most two values of for which. So there are at most two points of inflection. If the iscriminant of a quaratic is less than zero, there are no real solutions. Discriminant (6 ) 4ac 6 96 ac ( 8 ac) If 8ac then iscriminant < an so there are no solutions to. Therefore if 8 ac, then C has no points of inflection. Pearson Eucation Lt 7. Copying permitte for purchasing institution only. This material is not copyright free. 6