RI C8A Graphing Techniques Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________ Tutorial 8A: Graphing Techniques Page 1 of 4 Tutorial 8A: Graphing Techniques Section A (Basic Questions) 1 The equation of a curve is 32236 4yx x x  . Find ddyx and hence show that y is strictly increasing for all real values of x. [ 266 6xx ] Solution 2 2 2 2 d 66 6d 6( ) 6 1166 24 196 22 y xxx xx x x         Since 2211 96( ) 0 , 6( ) 022 2xx x x     Since d 0,dy xx    y is strictly increasing for all real values of x.(shown) 2 For each of the curves with the following equations, find ddyx and hence find the exact coordinates of the stationary points. Determin e the nature of each stationary point using the first or second derivative test. (a) y = x(ln x)2 , where x > 0, (b) 2 2 1 1 xxy xx  , [ (a) (1,0) min. (e2, 4e2) max (b) (–1, 3) max, (1, 13 ) min ] v Solution (a) y = x(ln x)2 , where x > 0  2 2 d1 2l n ( l n )d 2l n ( l n ) ln (2 ln ) y xx xxx xx xx     For stationary points: d 0dyx 

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _____________________________ Tutorial 8A: Graphing Techniques Page 2 of 4  2 ln (2 ln ) 0 ln 0 or 2 ln 0 1 or e xx xx xx      2 2 d1 1 ln (2 ln )d 22ln y xxxx x xxx    When 1x , 21(ln 1) 0y 2 2 1 d2 2 ln1 2 0d1 1xyx     so (1,0) a minimum point. When 2ex  , 22 2 2e( l n e) 4 ey   2 2 2 2 2 22 22 e d 2 24 22 ln e 0d e eeeexyx           so (e2, 4e2) is a maximum point. Alternatively, to verify the nature of the stationary points, we may use a GC, with screenshot as follows: (b) 2 2 2 2 22 2 22 22 dd 1 dd 1 d2 1d1 (1 ) ( 2 1 )2 (1 ) 2( 1) (1 ) 2( 1)( 1) (1 ) yx x xx x x x xx x xx x x xx x xx xx xx               For stationary points: d 0dyx 

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _____________________________ Tutorial 8A: Graphing Techniques Page 3 of 4 22 2( 1)( 1) 0(1 ) 10o r 10 1 or 1 xx xx xx xx       When 1x , 2 2 11 11 11 13y   so (1, 13 ) is a minimum turning point. When 1x  , 2 2 (1 ) 1 1 3(1 ) 1 1y    so ( 1, 3 ) is a maximum turning point. To verify the nature of these stationary points, we may use the GC to apply the 2nd derivative test, with screenshot as follows: Since 2 2 1 d 40d xyx    and 2 2 1d 0.444 0d xyx   , by 2nd derivative test, ( 1, 3 ) is a maximum turning point and (1, 13 ) is a minimum turning point. x 11 1 (1 ) (1 )xx ve 0 ve d d y x ve 0 ve Tangent Nature of Stationary Point Minimum turning point x (1 ) 1 (1 )  (1 ) (1 )xx ve 0 ve d d y x ve 0 ve Tangent Nature of Stationary Point Maximum turning point

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _____________________________ Tutorial 8A: Graphing Techniques Page 4 of 4 3 Sketch, on separate diagrams, the graphs of (a) 21 xy x  , (b) 621yx x  , giving the equations of any asymptotes and the coordinates of any points of intersection with the x- and y-axes. Solution (a)  2 1 11 3 1 31 1 xy x x x x        Asymptotes: 1y , 1x (b) 621yx x  Asymptotes: 21yx , 0x 