OTE: Algebraic expressions folloM FORTRAII conventions use full colculator preci
ID: 3878973 • Letter: O
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OTE: Algebraic expressions folloM FORTRAII conventions use full colculator precizion for intermediste values use the bisection method with the function defined by (x +2.51) 5.57 B(x3.06) 5.57 #1s( (A-1)..2). (1+24) (x) (P1 2.)(P2 (4.35)) Start ilith interval (x1eft.xright) (.2.51, 3.06) The function velues st these end points are (xleft) 2.38 The new approxinstion interval bracketing the root efter OlE bisection i: Function values at these - end point re and-. 735 The new approxinstion interval bracketing the root after ONE MORE bisection is ( _s Function values at these end points are: 1.75831 The new approxinstion interval bracketing the root If Xnid satisfies the convergence criterior If(xmid)IExplanation / Answer
After first bisection
Xmid= (-2.51+3.06)/2 =0.275
putting thise value in the expression F(X), we get -0.735
F(Xleft)=2.88
since F(Xleft)*F(Xmid) <0 the next end points will be (-2.51,0.275)
so the answer of the first part will be (d) [Selection for Blank Number 1 will be -2.51]
Selection for Blank number 2 will be (d) 0.275
Selection for Blank Number 3 will be (b) 2.88
Going to the next iteration,
Xmid= (-2.51+0.275)/2= -1.1175
putting this value in the expressions given above we get
F(Xmid)=1.75031
now,F(Xleft) * F(Xmid) >0
therefore next end point will be (-1.1175,0.275)
Selection for Blank no 4 will be (e) -1.1175
Selection ofr blank no 5 will be (e) 0.275
for blank number 6 we have to find the value of the expression f(X) at X= 0.275 which we have found earlier in this problem and that is -0.735
so the selection for blank no 6 will be (c) -0.735
for the next iteration .similarly we will follow the steps
Xmid=(-1.1175+0.275)/2 = -0.42125
F(Xmid)=-4.24124
F(Xleft) * F(Xmid) <0
therefore teh next end points will be (-1.1175,-0.42125)
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