Question

What is the solution of the equation given below by using bisection method upto four decimal places .(Consider the root lying on positive quadrant only and compute the root till five iteration only) f(x)=Xe^2x-3x^2-5

Answer 1

Given:

Equation f(x) = x$e^{2x}$ - 3x² - 5

To find:

Solution of the equation using bisection method upto four decimal places

Solution:

1. Given equaton is: f(x) = x$e^{2x}$ - 3x² - 5
2. The value of f(x) at a= 1 and b=2 is:
3. f(x = 1) = $e^{2} -8$ = -0.611 < 0
4. f(x = 2) = $2e^{4} - 12 -5$ = 92.2 >0
5. The given function is continuous and there is a root that lies in the interval [1, 2].

Iteration 1:

• Let “s1” be the midpoint of the interval.
• I.e., s1 = (1+2)/2
• s1 =3 / 2
• s1 = 1.5
• Therefore, the value of the function at “s1” is
• f(s1) = f(1.5) = (1.5)$e^{3}$ - 6.74 - 5 = 18.388 > 0
• The root lies between a=1 and a=1.5.

Iteration 2:

• Let “s2” be the midpoint of the interval.
• I.e., s2 = (1+1.5)/2
• s2 =1.25
• Therefore, the value of the function at “s2” is
• f(s2) = f(1.25) = (1.25)$e^{2*1.25}$ - 3×1.25² - 5 = 5.54 > 0
• The root lies between a=1 and a=1.25.

Iteration 3:

• Let “s3” be the midpoint of the interval.
• I.e., s3 = (1+1.25)/2
• s3 =1.125
• Therefore, the value of the function at “s3” is
• f(s3) = f(1.125) = (1.125)$e^{2*1.125}$ - 3×1.125² - 5 = 1.877 > 0
• The root lies between a=1 and a=1.125.

Iteration 4:

• Let “s4” be the midpoint of the interval.
• I.e., s4 = (1+1.125)/2
• s4 =1.0625
• Therefore, the value of the function at “s4” is
• f(s4) = f(1.0625) = (1.0625)$e^{2*1.0625}$ - 3×1.0625² - 5 = 0.51 > 0
• The root lies between a=1 and a=1.0625.

Iteration 5:

• Let “s5” be the midpoint of the interval.
• I.e., s5 = (1+1.0625)/2
• s5 =1.03125
• Therefore, the value of the function at “s5” is
• f(s5) = f(1.03125) = (1.03125)$e^{2*1.03125}$ - 3×1.03125² - 5 = -0.0790

Thus after the fifth iteration we get the solution of the equation as -0.0790 at x = 1.03125