1708but this is a graph of 2x = x3
here we have to draw the graph of 2x = x3+1
19jagaran's graph is right ....
@maverick.....look the graph is shifted by 1 unit along the x -axis ! that is done !
23no of solutions = 4
one intersection is between the x values of (-1,0) as seen from graph ,
other points of intersection are x =0 and x=1
now for x = 2 ,
LHS = 4 , RHS = 9
so LHS > RHS ...(1)
now for x = 100
LHS = 2100 = (24)25 = 1625
RHS = 1003+ 1 = 106 + 1 wich is < 1625
so RHS > LHS (2)
so from (1) and (2)
it is clear that at some value of x ....LHS became = RHS and then after that value of x, LHS always remains < RHS
so 4 points of intersections
1@mavarick just post ur graph to justify it
@ jagaran ur graph is pathetic
1good work â—‹|~_^|â—‹
but i was looking for a much better solution.......
something that strikes me at the time of the exam......
1what maverick???
do u think i am going to carry my laptop to examination hall so that wolfram can draw my graphs for me?????
23this thing striked me atleast ...btw instead of 100 u can take 10....
3we can have an idea abt roots by finding f'(x),f"(x) and f"'(x)
where f(x)=2x-(1+x3)
1
i m still getting 1 pt of intersection .hope graph is correct this time
23wat abt x = 0 ???
LHS = 20 = 1
RHS = 03 + 1 = 1 .....
in ur graph the functions dont intersect at x = 0
1@ JAGARAN
ur graphs still pathetic.........pls go over the fundamentals again
1agreed this is incorrrect. but is is possible to draw a graph as shown by maverick without using a graph plotter???????
i mean if u use fundamentals will u get a better graph.finding pts of intersection and then drawing a graph is not a part of fundamentals
23i dont think so we can find all points of intersections by using graph ..i mean for example if a question has has point of intersection at x =101.78
how can u find it by drawing a graph ??
u can find the no of intersections as i hav done above .....putting values in such functions isnt that tough ..lol i hav used x=100 to just elaborate it ..but u can even check it at x =10 ..