Use the Intermediate Value Theorem to show that there is a zero in between (2, 4
ID: 2878353 • Letter: U
Question
Use the Intermediate Value Theorem to show that there is a zero in between (2, 4). b) Use the Newton's method to find the zero of the equation f(x) = 0 correct to six decimal places. Use the definition of derivative f'(x) = lim_h rightarrow 0 f(x + h) - f(x)/h to find the equation of the normal line to the graph of f(x) = 2 - 5x - x^2 at P(-1, 1). If f(x) = |x|, show that f(-1) = f(1) but f'(c) notequalto 0 for every number c in the interval (-1.1). Why doesn't this contradict Roll's Theorem? Be very descriptive.Explanation / Answer
4)a) f(x)=2x-5-sinx
f(x) is continous in interval (2,4)
f(2)=2*2 -5 -sin2 =-1.909
f(4)=2*4 -5 -sin4 =3.7568
f(x) changes its sign from negative to positive ,
therefore by intermediate value theorem f(x) has atleast one zero in (2,4)
b)x1=2
f(x)=2x-5-sinx
f(2)=-1.9092974
f '(x)=2-cosx
f '(2)=2-cos2=2.4161468
x2=x1-f(x1)/f '(x1)
x2=2-(-1.9092974)/(2.4161468)
x2=2.79022409
x3=2.79022409-(0.23626510)/(2.93890256)
x3=2.70983180
x4=2.70983180-(0.00119289)/(2.90823029)
x4=2.7094216
x5=2.7094216-(0.000000031)/(2.9080586)
x5=2.709421
zero of f(x) is 2.709421
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