Question: Q 1 1 . The root o f equation c o s x = x e x i s t o b e found using

Q11. The root of equation cosx=xexistobe found using
Newton-Raphson iterations. The derivatives are given as
d(cosx)dx=-sinx,d(xex)dx=xex+ex
At iteration #0,x=1(the initial value). What is the value ofxat
iteration #2?Be accurate to five decimals points.
xk+1=xk-f(xk)f'(xk)
Q12. When you fit a linear function to the given data, what
will be the slope of the function?
[ni=1nxii=1nxii=1nxi2]{[a0],[ai]}={[i=1nyi],[i=1nxiyi]}
xy
14
45
56.5
Q 1 1 . The root o f equation c o s x = x e x i s

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