Connect 1-semester Access Card For Numerical Methods For Engineers
Connect 1-semester Access Card For Numerical Methods For Engineers
7th Edition
ISBN: 9781259168772
Author: Chapra, Steven, Canale, Raymond
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 13, Problem 1P

Given the formula

f ( x ) = x 2 + 8 x 12

(a) Determine the maximum and the corresponding value of x for this function analytically (i.e., using differentiation).

(b) Verify that Eq. (13.7) yields the same results based on initial guesses of x 0 = 0 , x 1 = 2 , and x 2 = 6 .

(a)

Expert Solution
Check Mark
To determine

To calculate: The maximum value of the function and the corresponding value of variable x for the function,

f(x)=x2+8x12

Answer to Problem 1P

Solution:

The maximum value of the function f(x)=x2+8x12 is 4 at x=4.

Explanation of Solution

Given Information:

The function f(x) is given as,

f(x)=x2+8x12

Calculation:

Evaluate the first order derivative of the function f(x).

f'(x)=2x+8 …… (1)

Equate equation (1) to zero and evaluate critical point.

2x+8=0x=82=4

Therefore, a critical point is at x=4.

Evaluate the second order derivative of the function f(x) to check if the function attains a maximum or minimum value at the critical point.

f''(x)=2 …… (2)

The sign of the second order derivative at x=4 is negative, therefore, at x=4 the given function attains a maximum value.

Substitute x=4 in the function f(x) to find the maximum value of the function.

f(x)=(4)2+8(4)12=16+3212=3228=4

Hence, the maximum value of the function at x=4 is 4.

(b)

Expert Solution
Check Mark
To determine

To prove: The results obtained in part (a) using the equation as follows:

x3=f(x0)(x12x22)+f(x1)(x22x02)+f(x2)(x02x12)2f(x0)(x1x2)+2f(x1)(x2x0)+2f(x2)(x0x1)

using the initial guess of x0=0,x1=2 and x2=6.

Explanation of Solution

Given Information:

The function is given as,

f(x)=x2+8x12

The initial guesses are x0=0,x1=2 and x2=6.

Formula used:

The expression for the parabolic interpolation is given as,

x3=f(x0)(x12x22)+f(x1)(x22x02)+f(x2)(x02x12)2f(x0)(x1x2)+2f(x1)(x2x0)+2f(x2)(x0x1)

Here, x0,x1and x2 are the initial guesses and f(x0),f(x1)and f(x2) are the values of the function at the respective initial guesses.

Proof:

Rewrite the equation of the parabolic interpolation:

x3=f(x0)(x12x22)+f(x1)(x22x02)+f(x2)(x02x12)2f(x0)(x1x2)+2f(x1)(x2x0)+2f(x2)(x0x1) …… (3)

Substitute x=0 in f(x) to get the value of f(x0).

f(x0)=(0)2+8(0)12=12

Thus, the value of f(x0) is 12.

Substitute x=2 in f(x) to get the value of f(x1).

f(x1)=(2)2+8(2)12=4+1612=0

Thus, the value of f(x1) is 0.

Substitute x=6 in f(x) to get the value of f(x2).

f(x2)=(6)2+8(6)12=36+4812=0

Thus, the value of f(x2) is 0.

Substitute 0 for x0, 2 for x2, 6 for x2, 12 for f(x0), 0 for f(x1) and 0 for f(x2) in equation (3).

x3=(12)[(2)2(6)2]+(0)[(6)2(0)2]+(0)[(0)2(2)2]2(12)[(2)(6)]+2(0)[(6)(0)]+2(0)[(0)(2)]=4

Thus, the value of x3 is 4, this value matches with the results obtained in part (a).

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