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

7/16

tep-by-step explanation:

Answer 2

Option B. 7/16

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