USe Newton-Raphson method to approximate the root of

f(x)=3x + sin (x) -e^x

Use initial guess x_{0 =2.0 and compute 3 iterations. for
each iteration, compute the relative percentage error .}

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GIVEN : f(x) = 3x + sin (x) - ex.  [ This gives : f'(x) = 3 + cos(x) - ex ] x0 = 2.0 n = 3 WE KNOW : Relative error at each step is given by :                           \epsilon_{\text {relative }}=\frac{\left|x_{i}-x_{i-1}\right|}{x_{i}} STEP 1 : x0 = 2.0  f(x0) = 3(2) + sin(2) - e2 = - 0.479759 f'(x0) = 3 + cos(2) - e2 = - 4.805203 x1 = x0 - f(x0) / f'(x0) = 1.900158 \epsilon_{\text {relative }}=\frac{\left|x_{i}-x_{i-1}\right|}{x_{i}}=\frac{|1.900158-2|}{1.900158}=0.052505 STEP 2 : x1 = 1.900158 f(x1) = 3(1.900158) + sin(1.900158) - e1.900158 = - 0.040230 f'(x1) = 3 + cos(1.900158) - e1.900158 = - 4.01039 x2 = x1 - f(x1) / f'(x1) = 1.890127 \epsilon_{\text {relative }}=\frac{\left|x_{i}-x_{i-1}\right|}{x_{i}}=\frac{|1.890127-1.900158|}{1.890127}=0.005307 STEP 3 : x2 = 1.890127 f(x2) = 3(1.890127) + sin(1.890127) - e1.890127 = - 0.000383 f'(x2) = 3 + cos(1.890127) - e1.890127 = - 3.93414 x3 = x2 - f(x2) / f'(x2) = 1.890030 <img/> ...