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Natali5045456 [20]
3 years ago
14

Question part points submissions used use newton's method with the specified initial approximation x1 to find x3, the third appr

oximation to the root of the given equation. (round your answer to four decimal places.) 2x3 − 3x2 + 2 = 0, x1 = −1
Mathematics
1 answer:
serious [3.7K]3 years ago
5 0

Set f(x)=2x^3-3x^2+2. Find the tangent line \ell_1(x) to f(x) at the point when x=x_1:

f'(x)=6x^2-6x\implies f'(x_1)=12 (slope of \ell_1)

\implies\ell_1(x)=12(x-x_1)+f(x_1)=12(x+1)-3=12x+9

Set x_2=-\dfrac9{12}, the root of \ell_1(x). The tangent line \ell_2(x) to f(x) at x=x_2 has slope and thus equation

f'(x_2)=\dfrac{63}8\implies\ell_2(x)=\dfrac{63}8\left(x+\dfrac9{12}\right)-\dfrac{17}{32}=7x+\dfrac{151}{32}

which has its root at x_3=-\dfrac{151}{224}\approx-0.6741.

(The actual value of this root is about -0.6777)

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<h3>Sin\theta_{1}=\frac{3\sqrt{21}}{17}</h3>

Solution given:

Cos\theta_{1}=\frac{10}{17}

\frac{adjacent}{hypotenuse}=\frac{10}{17}

equating corresponding value

we get

adjacent=10

hypotenuse=17

perpendicular=x

now

by using Pythagoras law

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substituting value

17²=x²+10²

17²-10²=x²

x²=17²-10²

x²=189

doing square root

\sqrt{x²}=\sqrt{189}

x=3\sqrt{21}

now

In I Quadrant sin angle is positive

Sin\theta_{1}=\frac{perpendicular}{hypotenuse}

<h3>Sin\theta_{1}=\frac{3\sqrt{21}}{17}</h3>
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