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
[tex]f(x)=0[/tex]
To find the approximate value of x, we use the following iterative formula
[tex]\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}[/tex]
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:
[tex]4^{-x}+5=3^x+4[/tex]
We construct a function f(x) which can be equated to zero and find its roots. Thus we define
[tex]f(x)=4^{-x}+5-3^x-4[/tex]
[tex]f(x)=4^{-x}+1-3^x[/tex]
To solve the original equation, it's the same as finding the roots of f(x)=0
The starting point [tex]x_1[/tex] will be obtained from the graph provided in the figure
[tex]x_1=0.5[/tex]
Lets find the derivative
[tex]f'(x)=-ln4\ 4^{-x}-ln3\ 3^x[/tex]
Evaluating in the point [tex]x_1=0.5[/tex]
[tex]f(0.5)=4^{-0.5}+1-3^{0.5}=-0.23205[/tex]
[tex]f'(0.5)=-ln4\ 4^{-0.5}-ln3\ 3^{0.5}=-2.596[/tex]
We find the next value:
[tex]\displaystyle x_{2}=0.5-\frac {-0.23205}{-2.596}[/tex]
[tex]x_2=0.4106[/tex]
Let's perform another iteration
Evaluating in the point [tex]x_2=0.4106[/tex]
[tex]f(0.4106)=4^{-0.4106}+1-3^{0.4106}=-0.00408[/tex]
[tex]f'(0.4106)=-ln4\ 4^{-0.4106}-ln3\ 3^{0.4106}=-2.50946[/tex]
We find the next value
[tex]\displaystyle x_{3}=0.4106-\frac {-0.00408}{-2.50946}[/tex]
[tex]x_3=0.409[/tex]
This is a very good approximation since
[tex]f(0.409)=-0.0000378281[/tex]
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
