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Determine the real roots of f(x)=-0.5x2+2.5x+4.5 1) Graphically 2) Using the quadratic formula 3) Using three iterations of the bisection method to determine the highest root. Employ initial guesses of xl=5 and xu=1. Compute the estimated error and the true error after each iteration.

1) A plot indicates that roots occur at about x = –1.4 and 6.4.

Quadratic formula

First iteration:

Third iteration

Determine the real root of f(x)=-26+85x-91x2+44x3-8x4+x5 1) Graphically 2) Using bisection to determine the root to εs=10%. Employ initial guesses of xl=0.5 and xu=1.0 3) Perform the same computation as in (2) but use the false-position method and εs=0.2%

A plot indicates that a single real root occurs at about x = 0.58.

Bisection:

First iteration

The process can be repeated until the approximate error falls below 10%. As summarized below, this occurs after 4 iterations yielding a root estimate of 0.59375.

False position: First iteration:

Second iteration:

The process can be repeated until the approximate error falls below 0.2%. As summarized below, this occurs after 4 iterations yielding a root estimate of 0.57956.

Determine the highest real root of f(x)=2x3-11.7x2+17.7x-5 1) Graphically 2) Newton-Raphson method (three iterations, x0 =3) 3) Secant method (three iterations, x1=3,x0=4)

Graphical

Newton-Raphson

Secant

Employ the Newton-Raphson method to determine a real root for f(x)=-2+6x-4x2+0.5x3 using initial guesses of 1) 4.52 2) 4.54 Discuss and use graphical and analytical methods to explain any peculiarities in your results

The Newton-Raphson method can be set up as

Using an initial guess of 4.52, this formula jumps around and eventually converges on the root at 0.214333 after 21 iterations:

The reason for the behavior is depicted in the following plot. As can be seen, the guess of x = 4.52 corresponds to a nearzero negative slope of the function. Hence, the first iteration shoots to a large negative value that is far from a root.

Using an initial guess of 4.54, this formula jumps around and eventually converges on the root at 6.305898 after 14 iterations:

The reason for the behavior is depicted in the following plot. As can be seen, the guess of x = 4.54 corresponds to a near-zero positive slope. Hence, the first iteration shoots to a large positive value that is far from a root.

Thanks~

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Determine the real roots of f(x)=-0.5x2+2.5x+4.5 1) Graphically 2) Using the quadratic formula 3) Using three iterations of the bisection method to determine the highest root. Employ initial guesses of xl=5 and xu=1. Compute the estimated error and the true error after each iteration.

1) A plot indicates that roots occur at about x = –1.4 and 6.4.

Quadratic formula

First iteration:

Third iteration

Determine the real root of f(x)=-26+85x-91x2+44x3-8x4+x5 1) Graphically 2) Using bisection to determine the root to εs=10%. Employ initial guesses of xl=0.5 and xu=1.0 3) Perform the same computation as in (2) but use the false-position method and εs=0.2%

A plot indicates that a single real root occurs at about x = 0.58.

Bisection:

First iteration

The process can be repeated until the approximate error falls below 10%. As summarized below, this occurs after 4 iterations yielding a root estimate of 0.59375.

False position: First iteration:

Second iteration:

The process can be repeated until the approximate error falls below 0.2%. As summarized below, this occurs after 4 iterations yielding a root estimate of 0.57956.

Determine the highest real root of f(x)=2x3-11.7x2+17.7x-5 1) Graphically 2) Newton-Raphson method (three iterations, x0 =3) 3) Secant method (three iterations, x1=3,x0=4)

Graphical

Newton-Raphson

Secant

Employ the Newton-Raphson method to determine a real root for f(x)=-2+6x-4x2+0.5x3 using initial guesses of 1) 4.52 2) 4.54 Discuss and use graphical and analytical methods to explain any peculiarities in your results

The Newton-Raphson method can be set up as

Using an initial guess of 4.52, this formula jumps around and eventually converges on the root at 0.214333 after 21 iterations:

The reason for the behavior is depicted in the following plot. As can be seen, the guess of x = 4.52 corresponds to a nearzero negative slope of the function. Hence, the first iteration shoots to a large negative value that is far from a root.

Using an initial guess of 4.54, this formula jumps around and eventually converges on the root at 6.305898 after 14 iterations:

The reason for the behavior is depicted in the following plot. As can be seen, the guess of x = 4.54 corresponds to a near-zero positive slope. Hence, the first iteration shoots to a large positive value that is far from a root.

Thanks~

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