the function has a real zero in the given interval. approximate this solution correct to two decimal places: f(x)=x^4-x^3-7x^2+5x+10; (2,3)
y =x^4-x^3-7x^2+5x+10
y' = 4x^3 -3x^2 -14x +5
start with x1 = 2.5
calculate y there call it y1
calculate y' there, call it m
next guess at x2 = x1 - y/m
thank you for your time and work. i really appreciate it. god bless you.
To approximate the real zero of a function in a given interval, you can use a numerical method called the bisection method. Here's how you can apply this method to approximate the solution to two decimal places for the given function f(x) = x^4 - x^3 - 7x^2 + 5x + 10 in the interval (2, 3):
Step 1: Check the signs
Evaluate f(2) and f(3) to determine if there is a real zero within the given interval.
f(2) = (2)^4 - (2)^3 - 7(2)^2 + 5(2) + 10 = 4 - 8 - 28 + 10 + 10 = -12
f(3) = (3)^4 - (3)^3 - 7(3)^2 + 5(3) + 10 = 81 - 27 - 63 + 15 + 10 = 16
Since f(2) is negative and f(3) is positive, there is a real zero within the interval (2, 3).
Step 2: Apply the bisection method
The bisection method starts by finding the midpoint of the interval and then checking the sign of the function at that point. If the function value is positive, the midpoint becomes the new upper bound of the interval; if the function value is negative, the midpoint becomes the new lower bound of the interval. This process is repeated until the interval becomes very small and the approximation is achieved.
Start with the interval (2, 3):
- Midpoint: x = (2 + 3) / 2 = 2.5
- Evaluate f(2.5):
f(2.5) = (2.5)^4 - (2.5)^3 - 7(2.5)^2 + 5(2.5) + 10 ≈ -0.2275 (rounded to four decimal places)
Since f(2.5) is negative, we update the lower bound of the interval to 2.5. Thus, the new interval is (2.5, 3).
Repeat the process:
- Midpoint: x = (2.5 + 3) / 2 = 2.75
- Evaluate f(2.75):
f(2.75) = (2.75)^4 - (2.75)^3 - 7(2.75)^2 + 5(2.75) + 10 ≈ 0.9607 (rounded to four decimal places)
Since f(2.75) is positive, we update the upper bound of the interval to 2.75. Thus, the new interval is (2.5, 2.75).
Repeat the process:
- Midpoint: x = (2.5 + 2.75) / 2 = 2.625
- Evaluate f(2.625):
f(2.625) = (2.625)^4 - (2.625)^3 - 7(2.625)^2 + 5(2.625) + 10 ≈ -0.1664 (rounded to four decimal places)
Since f(2.625) is negative, we update the lower bound of the interval to 2.625. Thus, the new interval is (2.625, 2.75).
Continue this process iteratively until the interval becomes very small, and the approximation will get closer to the real zero. In this case, the solution will be approximately 2.69 (rounded to two decimal places) after a few more iterations.
Note: The bisection method is an iterative process, and the number of iterations required to reach a desired level of accuracy can vary. The more iterations you perform, the more accurate the approximation will be.