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#3
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DarkCalculis Dec 7, 2017
#2
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+3

a)     g(x)  =  f(x + 1)

The domain of  f  is  (-1, 1) ,  so   x + 1   has to be in the interval   (-1, 1) .

-1  <  x + 1  <  1      Subtract  1  from each part of the inequality.

-2  <  x  <  0

So the domain of  g(x)  is  (-2, 0) .

b)     h(x)  =  f(x) + 1

The domain of  f  is  (-1, 1) ,  so   x   has to be in the interval   (-1, 1) .

So the domain of  g(x)  is  (-1, 1) .

c)     j(x)  =  f( 1/x )

The domain of  f  is  (-1, 1) ,  so   1/x   has to be in the interval   (-1, 1) .

-1  <  1/x  < 1

we can see that all  x  values  >  1  cause  1/x  to be within the desired range,

and all  x  values  <  -1  cause  1/x  to be within the desired range.

So the domain of  g(x)  is  (-∞, -1) U (1, ∞) .

d)     k(x)  =  f( √x )

The domain of  f  is  (-1, 1) ,  so   √x   has to be in the interval   (-1, 1) .

-1  <  √x  <  1       To solve this inequality, let's split it into two parts.

-1  <  √x               This is true for all non-negative  x  values, so...

x   ≥  0

and

√x  <  1

x  <  1

We can check this with a graph:  https://www.desmos.com/calculator/wqoa050mj0

So the domain of  k(x)  is  [0, 1) .

e)     $$l(x)=f(\frac{x+1}{x-1})$$

The domain of  f  is  (-1, 1) ,  so   $$\frac{x+1}{x-1}$$   has to be in the interval   (-1, 1) .

-1  < $$\frac{x+1}{x-1}$$  <  1

The easiest way to solve this inequality is, again, with a graph.

So the  x  values that cause  $$\frac{x+1}{x-1}$$  to be in the desired range are those  <  0 .

So the domain of  l(x)  is  (-∞, 0)

hectictar Dec 7, 2017

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