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a) Suppose the domain of $f$ is $(-1,1)$. Define the function $g$ by \(g(x)=f(x+1).\) What is the domain of $g$?

 

b) Suppose the domain of $f$ is $(-1,1)$. Define the function $h$ by \(h(x)=f(x)+1.\) What is the domain of $h$?

 

c) Suppose the domain of $f$ is $(-1,1)$. Define the function $j$ by \($j(x)=f(1/x).\) What is the domain of $j$?

 

d) Suppose the domain of $f$ is $(-1,1)$. Define the function $k$ by \(k(x)=f\left(\sqrt x\right)\) What is the domain of $k$?

 

e)Suppose the domain of $f$ is $(-1,1)$. Define the function $\ell$ by \(\ell(x)=f\left(\frac{x+1}{x-1}\right)\) What is the domain of $\ell$?

michaelcai  Dec 5, 2017

Best Answer 

 #2
avatar+7096 
+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

 

By looking at a graph,  here:  https://www.desmos.com/calculator/tswnd5ukqx  ,

 

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
 #1
avatar+598 
0

I really need help on these questions. Any help is appreciated!

michaelcai  Dec 7, 2017
 #2
avatar+7096 
+3
Best Answer

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

 

By looking at a graph,  here:  https://www.desmos.com/calculator/tswnd5ukqx  ,

 

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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