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It is an attracting point, especially due to the fact that I used the number 1 as my first “guess” which made every other answer to the iteration “-1” because it is one of the numbers that solve the problem evenly. iii. Iterations on page 8 It is a repelling point, and maybe this is because the first “guess” number was a decimal, but also, the staircase here is found to be on the left side of the graph, whereas the above iteration was a right staircase. B. Consider the equation x = -0. 5x + 2.

It is an attracting point, staircasing left of the fixed point. Again, here I noticed that I used a whole number as a first “guess” to be multiplied by a fraction of one in the equation. I am beginning to draw that if the first “guess” is also a decimal or fractional answer that is to be equated, that may affect the attracting or repelling nature of these points. C. Consider the equation x = -2x – 4 Iterations on page 8 It is a repelling point when staircasing to the right of the fixed point -4/3.

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These iterations seem to jump all over the graph here, switching from large negative to positive answers. Maybe this still has to do with the fact that here we are equating only whole numbers… D. Consider the equation (x + 1)(x – 0. 5)(x – 1) = 0 Iterations on page 9 It is a repelling fixed point in the consideration of x = -1 for the first fixed point and x = -1. 5 for the first “guess. ” Once more, I have a feeling that the fractional and whole numbers play a role in attraction.

It is an attracting fixed point in light of the x = 0. 5 and the first “guess” being x = 0. 9. It is a repelling fixed point when examining the last graph where x = 1 and the first “guess” is x = 1. 1. Once looking over these last equations, I am beginning to think that a greater meaning lies behind why some equations have repelling and some have attracting fixed points. E. Table of Attracting Fixed Point and Repelling Fixed Points Tables on page 10.

Conjecture: The condition, which must be present for a fixed point to be attracting, is apparent to me that every one has a fractional slope of the tangent, or f ‘ (x). The repelling points also have a few fractional tangential slopes, but all of the attracting ones do as a trend. ii. Testing the conjecture: Unfortunately, this conjecture did not hold true for the equation f(x) = x^5 -2 when viewing its staircase from the left, as its fixed point seems to be centered around -3, and the tangential slope of that becomes a positive term, -405. (Iterations on page 10. )