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Mathematics 19 Online
OpenStudy (neo92):

I want to find the roots of x^3+2x^2-3x-1=0 using bisection method. But Im not sure how to select the initial approximation. Can you tell me how select a suitable initial approximation to this?

OpenStudy (anonymous):

i wl tell u d trick

OpenStudy (neo92):

ok. tl me.. my buk menyons something called taylor method. But i hav no ida abt it.

OpenStudy (neo92):

mentions

OpenStudy (anonymous):

frst select the interval where sign of the equation changes

OpenStudy (anonymous):

exactly you have to choose smallest interval of two points say a and b such that f(a)<0 and f(b)>0

OpenStudy (anonymous):

absolutely right

OpenStudy (anonymous):

so that interval is (a,b)

OpenStudy (neo92):

can u tell me how to select the intervals?

OpenStudy (anonymous):

first method is hit n trial..................

OpenStudy (neo92):

can i differentiate and find it?

OpenStudy (anonymous):

gd there u cn find the points where slope is zero

OpenStudy (anonymous):

by differntiating u gt x= 0.58 n - 2.58

OpenStudy (neo92):

yep. so what should i do after that?

OpenStudy (anonymous):

then take points nearer to one point u wl see sign of equation changes similarly repeat for other point that u hav find after differntiation....

OpenStudy (neo92):

|dw:1358689959945:dw|

OpenStudy (neo92):

but does the sign of the eqn changes on a occassion like this?

OpenStudy (anonymous):

i didn't get wat r u saying????

OpenStudy (neo92):

You said I should select tha interval where the sign of the equation changes.

OpenStudy (anonymous):

w8 i m clearing

OpenStudy (anonymous):

first put x = 0 in the original cubic equation tl me wat u get ?

OpenStudy (anonymous):

asap

OpenStudy (neo92):

1

OpenStudy (anonymous):

dear it's negative keep all the expression on the left while putting so u gt -1

OpenStudy (anonymous):

then put x = 1 u wl get -1 again .........till now sign of the function is negative kk

OpenStudy (neo92):

yep.. i got it

OpenStudy (anonymous):

nw put x = 2 u wl gt 9 which is positive so ur initial approximation is (1,2)

OpenStudy (anonymous):

hope u gt it dear NEO

OpenStudy (anonymous):

THEN APPLY BISECTION

OpenStudy (neo92):

Okey.. i gt it now. so is that the only way.. ?

OpenStudy (anonymous):

yup dear n thanks fr awarding me the medal

OpenStudy (neo92):

:)

OpenStudy (neo92):

so I hav to apply the same method for an eqn like \[3x-e ^{x}=0\]

OpenStudy (anonymous):

u can dear........

OpenStudy (anonymous):

here interval is wat tl me asap?

OpenStudy (neo92):

wait a min

OpenStudy (anonymous):

i wl

OpenStudy (neo92):

is it (1,2) ?

OpenStudy (anonymous):

w8

OpenStudy (anonymous):

by putting 1 u get 1.74 n fr x = 0 u gt -1 so changes here first interval is (0,1)

OpenStudy (neo92):

oh yes.. i hav missed the -sign when calculating and i think 1 gives 0.28..... (3-e)

OpenStudy (anonymous):

gt it dear

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