2) Consider the function f(x) = 30x - 31x + 10x -1. The roots of f(r)...

 

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2) Consider the function f(x) = 30x³ - 31x² + 10x -1. The roots of f(r) are,, and . (a) Starting with the interval [0, 0.9], use the Bisection Method (at least 5 iterations) to search for a root of f. Which root of f does it look like you will find? (b) Starting with the interval [0, 0.6], use the Bisection Method (at least 5 iterations) to search for a root of f. Which root of f does it look like you will find? Notice that the intervals [0, 0.9] and [0, 0.6] both contain all three roots of f, but the Bisection Method will only find one root at a time (which one depends on your starting interval). (3) Use the Intermediate Value Theorem to explain why every positive real number r must have a square root. That is, there is some real number z such that 2² = r. [Hint: The fact that lim ² = co means that r² must eventually be larger than any particular value.] 2) Consider the function f(x) = 30x³ - 31x² + 10x -1. The roots of f(r) are,, and . (a) Starting with the interval [0, 0.9], use the Bisection Method (at least 5 iterations) to search for a root of f. Which root of f does it look like you will find? (b) Starting with the interval [0, 0.6], use the Bisection Method (at least 5 iterations) to search for a root of f. Which root of f does it look like you will find? Notice that the intervals [0, 0.9] and [0, 0.6] both contain all three roots of f, but the Bisection Method will only find one root at a time (which one depends on your starting interval). (3) Use the Intermediate Value Theorem to explain why every positive real number r must have a square root. That is, there is some real number z such that 2² = r. [Hint: The fact that lim ² = co means that r² must eventually be larger than any particular value.]

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2a Starting with the interval 0 09 using the Bisection method we can find the root of fx 3033110x1 T... View the full answer