Question

given f:x --> x^3 - 3x + 2 and g:x --> x-1 both X is real number.
(a) Find fg(x) and hence determine the starionary point.
(b) Sketch the graph of y=h(x)

Answer 1

fg(x)=x^3-3x^2+4

Answer 2

stationary pt=> dy/dx =0

Answer 3

(fg)'(x)=3x^2-6x

Answer 4

0=3x^2-6x

Answer 5

0=x(3x-6)

x=0 or x=2

Answer 6

when
x=0, fg(0)=4
x=2,fg(2)=0

Answer 7

for sure this graph hv no asymptote so no need to calc limit

Answer 8

but u need to find the root

Answer 9

root=>fg(x)=0

Answer 10

x^3-3x^2+4=0

Answer 11

err w8
back at that dy/dx = 0 tat one nid do the differentiation step rite?

Answer 12

direct differentiation u can see the step without writing down

Answer 13

w8 a i see back

Answer 14

ok
all above understand so for this qns roots also included in the marks rite?

Answer 15

for stpm, u can straight away pressing the calc to get the ans coz in this chapter they didnt ask u polynomial

Answer 16

they ask function

Answer 17

so u mean jus tat x^3-3x^2+4=0 then str8 roots ok d?

Answer 18

if the question is ask abt polynomial u need to show the step

Answer 19

oo
kk so part (a) consider done d rite
?

Answer 20

ya

Answer 21

bt u need to wright the conclusion for stationary pt (0,4) and (2,0)

Answer 22

write*

Answer 23

actually what does stationary pt means

Answer 24

so u can press calc to get the root

x=-1 or x=2

Answer 25

max or min pt

Answer 26

oh lol zzzz i got graph d but duno wad de hell is tat stationary pt nia lol
new term

Answer 27

bt u need to calc the nature for the stationary pt for part b

Answer 28

so u need to calc d^2x/dy^2

Answer 29

(fg)'(x)=3x^2-6x
(fg)''(x)=6x-6

Answer 30

(fg)''(x)=6x-6
for pt (0,4)
(fg)''(0)=6(0)-6=-6<0 is max pt
for pt (2,0)
(fg)''(2)=6(2)-6=6>0 is min pt

Answer 31

wait is d^2y/dx^2 rite?

Answer 32

second derivative

Answer 33

diff twice

Answer 34

d term "nature" means the pt weather is max or min or pt of inflection

Answer 35

pt of inflection means (fg)"(x)=0

Answer 36

since u hv all infos, u can draw the graph d