Need help with functions

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April 26th 2010, 08:23 AM
NewtoMath

Need help with functions

Need to be able to do all these (and then some) before exam tomorrow, any help whatsoever would be appreciated.

1. State domain and range of f(x) 2-|x-1|

2. f(x) =
{√(x-1) where x ≥ 1
{4 where x<1

Find f(a-1) in terms of a

2. Let g;[b,2] -> R where g(x) = 1 - x^2 , if b is the smallest real value of such that g has an inverse function, find b and inverse of g(x)

3. let f:S->R. f(x) = square root (4-x^20) and S be the set of all real values for x for which f(x) is defined. Let g:R->R, where g(x) = x^2 + 1

Find S and range of f and g.

4. Let 'a' be a positive number, let f:[2, infinite) -> R, f(x) = a - x and let g:(-infinite, 1] ->R, g(x) = x^2 + a. Find all values of 'a' for which f(g(x)) and g(f(x)) are defined.

5. find implied domain of square root (x^2 -x-2) and implied domain for square root (x-3x^2)

6.Sketch this hybrid function (just want to know how to go about doing this)
f(x)=
{2x+6 where 0<2x≤2
{-x + 5 where =4≤x≤0
{=4 where x < -4
April 28th 2010, 11:48 PM
Amer

Quote:

Originally Posted by NewtoMath

Need to be able to do all these (and then some) before exam tomorrow, any help whatsoever would be appreciated.

1. State domain and range of f(x) 2-|x-1|

2. f(x) =
{√(x-1) where x ≥ 1
{4 where x<1

Find f(a-1) in terms of a

2. Let g;[b,2] -> R where g(x) = 1 - x^2 , if b is the smallest real value of such that g has an inverse function, find b and inverse of g(x)

3. let f:S->R. f(x) = square root (4-x^20) and S be the set of all real values for x for which f(x) is defined. Let g:R->R, where g(x) = x^2 + 1

Find S and range of f and g.

4. Let 'a' be a positive number, let f:[2, infinite) -> R, f(x) = a - x and let g:(-infinite, 1] ->R, g(x) = x^2 + a. Find all values of 'a' for which f(g(x)) and g(f(x)) are defined.

5. find implied domain of square root (x^2 -x-2) and implied domain for square root (x-3x^2)

6.Sketch this hybrid function (just want to know how to go about doing this)
f(x)=
{2x+6 where 0<2x≤2
{-x + 5 where =4≤x≤0
{=4 where x < -4

do not post more than two question per thread

I will solve first one

$f(x) = 2 - \mid x-1 \mid$

domain all real numbers, range $(-\infty , 2]$

2)
$f(x) = \left\{\begin {array}{cc} \sqrt{x-1} & x\geq 1 \\ 4& x<1 \end{array} \right.$

$f(a-1) = \left\{\begin {array}{cc} \sqrt{a-2} & a-1\geq 1 \\ 4& a-1<1 \end{array} \right.$

3) b=0 since y is symmetric about y-axis so if the domain exceed 0 from the left the function will not be one-one and you can find inverse function for one-one functions just,
first let determine the range of our function [-3,1]
the inverse will be

$y =1- x^2 \Rightarrow x = \sqrt{1-y}$

g:[-3,1]-->[0,2]

$g(x) = \sqrt{1-x}$
April 29th 2010, 11:25 AM
stapel

6) To learn how to graph piecewise functions, try an online lesson or two. (Wink)