How do you find $f^{- 1} x$ $f^-1x$ and how must x be restricted in $f^{- 1} x$ $f^-1x$ given $f (x) = 4 + 2 cos (x - 3)$ $f(x)=4+2cos(x-3)$ and $3 \leq x \leq 3 + π$ $3<=x<=3+pi$ ?

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How do you find $f^{- 1} x$ $f^-1x$ and how must x be restricted in $f^{- 1} x$ $f^-1x$ given $f (x) = 4 + 2 cos (x - 3)$ $f(x)=4+2cos(x-3)$ and $3 \leq x \leq 3 + π$ $3<=x<=3+pi$ ?

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Answer 1 · 2018-01-31T11:31:55

As with all inverse functions, start by taking equation $y = f (x)$ $y=f(x)$ and substituting $x = f (y)$ $x=f(y)$

We have $y = f (x) = 4 + 2 cos (x - 3)$ $y = f(x) = 4 +2cos(x-3)$ , hence we let: $x = 4 + 2 cos (y - 3)$ $x = 4 + 2cos(y-3)$

Doing some algebraic manipulation we get:

$\frac{1}{2} (x - 4) = cos (y - 3)$ $1/2(x-4) = cos(y-3)$
$:. y -3 = cos^-1(1/2(x-4))$
$:. y = cos^-1(1/2(x-4))+3$

This is our $f^-1$ , but first we must find the new domain.

Since $dom f = [3, 3+pi]$ and the cosine function is one-to-one within any interval of $pi$ , there is no need to restrict further. Now $domf$ becomes $ranf^-1$ .

The implied domain of $f^-1$ gives $1/2(x-4) in [0,pi]$ , which implies that $x in [4,2pi +4]$

Hence:

$f^-1: [4,2pi +4] rarr RR, f^-1(x) = cos^-1(1/2(x-4))+3$

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How do you find $f^{- 1} x$ $f^-1x$ and how must x be restricted in $f^{- 1} x$ $f^-1x$ given $f (x) = 4 + 2 cos (x - 3)$ $f(x)=4+2cos(x-3)$ and $3 \leq x \leq 3 + π$ $3<=x<=3+pi$ ?

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How do you find $f^{- 1} x$ $f^-1x$ and how must x be restricted in $f^{- 1} x$ $f^-1x$ given $f (x) = 4 + 2 cos (x - 3)$ $f(x)=4+2cos(x-3)$ and $3 \leq x \leq 3 + π$ $3<=x<=3+pi$ ?