How do we approximate $f^{- 1} (x)$ $f^(-1)(x)$ ?

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How do we approximate $f^{- 1} (x)$ $f^(-1)(x)$ ?

If $f (x) = x + sin (x π)$ $f(x)=x+sin(xpi)$ , how could we go about trying to find its inverse over an interval on which $f$ $f$ is invertible?

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Answer 1 · 2017-11-17T22:44:39

We have:

$f (x) = x + sin (π x)$ $f(x)=x+sin(pix)$

If we let:

$y = x + sin (π x)$ $y = x+sin(pix)$ ..... [A]

Then, the inverse, $f^{- 1} (x)$ $f^(-1)(x)$ would normally be obtained by re-arranging equation [A] into the form:

$x = g (y)$ $x = g(y)$

Making $g$ $g$ the inverse.

However we cannot perform such a re-arrangement using the standard elementary functions that we are familiar with.

This is where the "approximation" component comes in because given any particular constant value of $x$ $x$ , (let us call this value $α$ $alpha$ , say) we would find the inverse $f^{- 1} (α)$ $f^(-1)(alpha)$ by solving numerically the equation:

$x + sin (π x) = α$ $x+sin(pix) =alpha$

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How do we approximate $f^{- 1} (x)$ $f^(-1)(x)$ ?

If $f (x) = x + sin (x π)$ $f(x)=x+sin(xpi)$ , how could we go about trying to find its inverse over an interval on which $f$ $f$ is invertible?

1 Answer

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How do we approximate f^(-1)(x)f−1(x)f^(-1)(x)?

Iff(x)=x+sin(xpi)f(x)=x+sin(xπ)f(x)=x+sin(xpi), how could we go about trying to find its inverse over an interval on which fff is invertible?

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How do we approximate $f^{- 1} (x)$ $f^(-1)(x)$ ?

If $f (x) = x + sin (x π)$ $f(x)=x+sin(xpi)$ , how could we go about trying to find its inverse over an interval on which $f$ $f$ is invertible?