Use the intermediate value theorem to find the real zero of the given polynomial correct to two decimal places.
[tex]f(x)=x^3+3 x-3[/tex]
The real zero correct to two decimal places is $\square$ .
(Round the final answer to two decimal places as needed. Round all intermediate values to four decimal places as needed.)
Answer (2)
Use the Intermediate Value Theorem to find an interval [ a , b ] where f ( a ) and f ( b ) have opposite signs.
Refine the interval by repeatedly evaluating f ( x ) at the midpoint c = ( a + b ) /2 and updating the interval.
Continue the iterations until the interval is sufficiently small to achieve the desired accuracy.
Approximate the root by the midpoint of the final interval and round to two decimal places: 0.82 .
Explanation
Understanding the Problem and the Intermediate Value Theorem We are asked to find the real zero of the polynomial f ( x ) = x 3 + 3 x − 3 using the Intermediate Value Theorem, correct to two decimal places. The Intermediate Value Theorem states that if a continuous function f has values f ( a ) and f ( b ) with opposite signs, then there exists at least one root between a and b .
Finding an Initial Interval First, we need to find an interval [ a , b ] such that f ( a ) and f ( b ) have opposite signs. Let's try a = 0 and b = 2 .
f ( 0 ) = 0 3 + 3 ( 0 ) − 3 = − 3 f ( 2 ) = 2 3 + 3 ( 2 ) − 3 = 8 + 6 − 3 = 11
Since f ( 0 ) = − 3 < 0 and 0"> f ( 2 ) = 11 > 0 , there is a root between 0 and 2 .
Refining the Interval Now, we will refine the interval using the bisection method. We repeatedly evaluate the function at the midpoint c = ( a + b ) /2 and update the interval accordingly.
Iteration 1: c = ( 0 + 2 ) /2 = 1 0"> f ( 1 ) = 1 3 + 3 ( 1 ) − 3 = 1 + 3 − 3 = 1 > 0 Since f ( 0 ) < 0 and 0"> f ( 1 ) > 0 , the root lies between 0 and 1 . Update the interval to [ 0 , 1 ] .
Iteration 2: c = ( 0 + 1 ) /2 = 0.5 f ( 0.5 ) = ( 0.5 ) 3 + 3 ( 0.5 ) − 3 = 0.125 + 1.5 − 3 = − 1.375 < 0 Since f ( 0.5 ) < 0 and 0"> f ( 1 ) > 0 , the root lies between 0.5 and 1 . Update the interval to [ 0.5 , 1 ] .
Iteration 3: c = ( 0.5 + 1 ) /2 = 0.75 f ( 0.75 ) = ( 0.75 ) 3 + 3 ( 0.75 ) − 3 = 0.421875 + 2.25 − 3 = − 0.328125 < 0 Since f ( 0.75 ) < 0 and 0"> f ( 1 ) > 0 , the root lies between 0.75 and 1 . Update the interval to [ 0.75 , 1 ] .
Iteration 4: c = ( 0.75 + 1 ) /2 = 0.875 0"> f ( 0.875 ) = ( 0.875 ) 3 + 3 ( 0.875 ) − 3 = 0.669921875 + 2.625 − 3 = 0.294921875 > 0 Since f ( 0.75 ) < 0 and 0"> f ( 0.875 ) > 0 , the root lies between 0.75 and 0.875 . Update the interval to [ 0.75 , 0.875 ] .
Iteration 5: c = ( 0.75 + 0.875 ) /2 = 0.8125 f ( 0.8125 ) = ( 0.8125 ) 3 + 3 ( 0.8125 ) − 3 = 0.5359802246 + 2.4375 − 3 = − 0.0265197754 < 0 Since f ( 0.8125 ) < 0 and 0"> f ( 0.875 ) > 0 , the root lies between 0.8125 and 0.875 . Update the interval to [ 0.8125 , 0.875 ] .
Iteration 6: c = ( 0.8125 + 0.875 ) /2 = 0.84375 0"> f ( 0.84375 ) = ( 0.84375 ) 3 + 3 ( 0.84375 ) − 3 = 0.601348877 + 2.53125 − 3 = 0.132598877 > 0 Since f ( 0.8125 ) < 0 and 0"> f ( 0.84375 ) > 0 , the root lies between 0.8125 and 0.84375 . Update the interval to [ 0.8125 , 0.84375 ] .
Iteration 7: c = ( 0.8125 + 0.84375 ) /2 = 0.828125 0"> f ( 0.828125 ) = ( 0.828125 ) 3 + 3 ( 0.828125 ) − 3 = 0.5685 + 2.484375 − 3 = 0.052875 > 0 Since f ( 0.8125 ) < 0 and 0"> f ( 0.828125 ) > 0 , the root lies between 0.8125 and 0.828125 . Update the interval to [ 0.8125 , 0.828125 ] .
Iteration 8: c = ( 0.8125 + 0.828125 ) /2 = 0.8203125 $f(0.8203125) = (0.8203125)^3 + 3(0.8203125) - 3 = 0.5520 + 2.4609375 - 3 = -0.0
After several iterations (as shown in the python calculation tool), we find that the root is approximately 0.8177. Rounding to two decimal places, we get 0.82.
Final Answer The real zero of the polynomial f ( x ) = x 3 + 3 x − 3 , correct to two decimal places, is approximately 0.82.
Examples
The Intermediate Value Theorem is used in various fields, such as engineering and computer graphics. For example, in engineering, it can be used to verify the stability of a system by checking if a function representing the system's behavior crosses a critical threshold. In computer graphics, it can be used to find the intersection points of curves, which is essential for rendering realistic images. Understanding the Intermediate Value Theorem helps in solving real-world problems where finding the roots of a function is crucial.
By applying the Intermediate Value Theorem to the polynomial f ( x ) = x 3 + 3 x − 3 , we narrow down the interval containing the root through iterative evaluations and find the real zero to be approximately 0.82 .
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