Question

Consider functions f and g.

What is the approximate solution after three iterations of successive approximations? Use the graph as a starting point.
(IN SCREENSHOT)

(I tried to solve it but I don't know where I went wrong, my procedure is in the second screenshot)

Answer 1

It depends on what method you're using to approximate. Newton's method is a popular choice. Let
`h(x) = g(x) - f(x) = 3\log_(10)(x-2) - \log_(10)(x)`, with derivative

`h'(x) = \frac1{\ln(10)} \left((3)/(x-2) - \frac1x\right)`

which follows from

`(d)/(dx) \log_(10)(x) = (d)/(dx) (\ln(x))/(\ln(10)) = \frac1{\ln(10)\,x}`

Judging by the plots, we have
`f(x)=g(x)` for some
`x` between 3 and 4. So let's take an initial approximation of
`x_0 = 3`. Newton's method involves taking the tangent line approximation to
`f(x)` at
`x=x_0`, then using the
`x`-intercept of this tangent line as the next approximation,
`x_1`.

The linear approximation at
`x_0=3` is

`h(x) \approx h(x_0) + h'(x_0) (x - x_0) \approx 1.15812x - 3.95148`

with intercept at
`x_1 \approx 3.41198`.

Repeat until you reach the desired threshold of accuracy. The next linear approximation at
`x_1` is

`h(x) \approx h(x_1) + h'(x_1) (x - x_1) \approx 0.79545 x-2.79758`

with intercept
`x_2 \approx 3.51698`.

The next approximation at
`x_2` is

`h(x) \approx h(x_2) + h'(x_2) (x - x_2) \approx 0.735382 x-2.58956`

with intercept
`x_3 \approx 3.52137`.

And so on. The actual solution has an approximate value of about

3.521379706804567569604081

so after just 3 approximations we're already less than
`10^(-5)` away.