Numerical Method (Bisection method) problem

[hashin03]

01 The function f is defined by
f(x) =x^2 + e^x -4

1. By drawing suitable sketches, show that f(x) has one negative root and one positive root.
2. Find the negative root of f(x) with a suitable starting point, correct to 3 decimal places.
3. Try to find the positive root of f(x).

02 (a) Sketch on the same diagram the curves y=2^(-x) and y=x^2.
(b) One of the points of intersection of these graphs has a positive x-coordinate. Given
this x-coordinate is a, show that 0<a<1.
(c) Find a correct to 2 decimal places and describe briefly how the iteration process
converges to a.



Pls help me with these questions. Stuck in them for a week. I don't get the concept of the numerical method as I've never done math with no exact solution. So frustrating. Is anyone familiar with these questions?

 

[HallsofIvy]

First, have you at least graphed [math]y= 2^{-x}[/math] and [math]y= x^2[/math]? If so, do you see that x= -2 is an exact solution and that there is another solution between 0 and 1?

I would start by writing [math]F(x)= 2^{-x}- x^2[/math]. Then [math]F(0)= 2^0- 0= 1[/math] and [math]F(1)= 2^{-1}- 1^2= \frac{1}{2}- 1= -\frac{1}{2}[/math]. The fact that the continuous function, F, is positive at x= 0 and negative at x= 1 tells us that it must be 0 at some point between 0 and 1. We don't know exactly but the "bisection" method tells us to try half way between, at x= 0. [math]F(1/2)= 2^{-1/2}- (1/2)^2[/math] which is, approximately, [math]0.707- 0.25= 0.457[/math]. Comparing signs, F(0.5) is positive and F(1) is negative so F must be 0 at some point between 0.5 and 1. Again, the "bisection method" says, try half way between which is x= 0.75. [math]F(0.75)= 2^{-3/4}- (3/4)^2[/math] which is, approximately, 0.5625. That is positive so there must be a solution between 0.75 and 1. Half way between is x= 0.825.

Continue like that. Calculate F(0.825) to see if it is negative or positive. If it is negative, there must be a solution between 0.5 and 0.825. If it is positive there must be a solution between 0.826 and 1. Which ever it is, take x half way between and continue until you have the required accuracy, two decimal places.

 

[Steven G]

I am thinking of a number between 0 and 128.
You guess 64 (the middle of 0 and 128).I tell you my number is lower than 64.
Then you guess 32 (the middle of 0 and 64).
I tell you my number is larger than 32.
You guess 48 (between 32 and 64)
....

Just do the same thing with your problem. The only difference is that I will not tell you higher or lower. You'll have to figure that out on your own.