Question

40 pts help! Approximate the solution to the equation above using three iterations of successive approximation. Use the graph below as a starting point.

Answer 1

Answer:

Option B. 7/16

Step-by-step explanation:

Approximation of Roots of Equations

Newton-Raphson's method is widely used to calculate approximate values of the roots of equations which cannot be solved with algebraic methods.

Let's suposse we need to solve the equation

$f(x)=0$

To find the approximate value of x, we use the following iterative formula

$\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}$

where f' is the first derivative of f

Each value will determine the next one which should be closer to the solution f(x)=0. We need an initial value to start the iterations

We want to solve this equation:

$4^{-x}+5=3^x+4$

We construct a function f(x) which can be equated to zero and find its roots. Thus we define

$f(x)=4^{-x}+5-3^x-4$

$f(x)=4^{-x}+1-3^x$

To solve the original equation, it's the same as finding the roots of f(x)=0

The starting point $x_1$ will be obtained from the graph provided in the figure

$x_1=0.5$

Lets find the derivative

$f'(x)=-ln4\ 4^{-x}-ln3\ 3^x$

Evaluating in the point $x_1=0.5$

$f(0.5)=4^{-0.5}+1-3^{0.5}=-0.23205$

$f'(0.5)=-ln4\ 4^{-0.5}-ln3\ 3^{0.5}=-2.596$

We find the next value:

$\displaystyle x_{2}=0.5-\frac {-0.23205}{-2.596}$

$x_2=0.4106$

Let's perform another iteration

Evaluating in the point $x_2=0.4106$

$f(0.4106)=4^{-0.4106}+1-3^{0.4106}=-0.00408$

$f'(0.4106)=-ln4\ 4^{-0.4106}-ln3\ 3^{0.4106}=-2.50946$

We find the next value

$\displaystyle x_{3}=0.4106-\frac {-0.00408}{-2.50946}$

$x_3=0.409$

This is a very good approximation since

$f(0.409)=-0.0000378281$

We need to pick one of the options and find none of them is close enough to 0.409. So we'll just choose the closest

Option B. 7/16=0.4375

Answer 2

Answer:

mhgjhg

Step-by-step explanation: