number of solutions

Ans: = 4

but this is a graph of 2^x = x³
here we have to draw the graph of 2^x = x³+1

jagaran's graph is right ....
@maverick.....look the graph is shifted by 1 unit along the x -axis ! that is done !

no 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

good work â—‹|~_^|â—‹
but i was looking for a much better solution.......
something that strikes me at the time of the exam......

http://www.wolframalpha.com/input/?i=(2^x)%3D(x^3)%2B1

what maverick???
do u think i am going to carry my laptop to examination hall so that wolfram can draw my graphs for me?????

this thing striked me atleast ...btw instead of 100 u can take 10....

we can have an idea abt roots by finding f'(x),f"(x) and f"'(x)

where f(x)=2^x-(1+x³)

agreed 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