Algebra Problems (joint variation etc..)?

Question

(1) Given: f(x)=3x-5 and g(x)=-5x+2

1. Find f[g(-1)]

2. Find g[f(4)]

(2)

Given f(x)=3x+4

1. Is f(x) a function? Why or why not?

2. Find f -1(x).

Is f -1(x) a function? Why or why not?

(3)

Given the relation R={(1,2),(-3,4),(0,-4),(1,0)}

1. Is R a function? Why or why not?

2. Which point lies on the y-axis?

3. Which point lies in Quadrant II?

4. Make a chart of the Relation

Answer 1

1) g(-1) = 5(-1)+2 =-5+2=-3

f(g(-1))=f(-3)=3(-3)-5 =-9 -5 =-14

(2)

1 ) yes, it is a function

For every element x in the domain, the function assigns a value in the range.

2) f^-1(x)

let y=3x+4

x=3y+4

x-4=3y

(x-4)/3=y

f^-(x) = (x-4)/3

Yes, f^-1(x) is a function (reasons are same as before)

3)1) No, it is not

For element 1 in the domain, it shows 2 values in the range namely 2 and 0

(1,2)(1,0) -- This is not allowed. You either show 2 or show 0, not both.

2)4 and -4 on the y-axis

3) (-3,4) x is negative and y is positive

Answer 2

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Answer 3

1) Given: f(x)=3x-5 and g(x)=-5x+2

1. f[g(-1)] = 3g(-1) - 5 = 3[-5(-1) + 2] - 5 = 16

2. g[f(4)] = -5f(4) + 2 = -5(3*4 - 5) + 2 = -33

2) Given f(x)=3x+4

1. f(x) is a function, because to every value of x there corresponds a value of f(x), and that value of f(x) is unique.

2. y = 3x + 4 --> x = (y - 4) / 3

f^-1(x) = (x - 4) / 3

f^-1 (x) is a function for the same reasons as above

3) Given the relation R={(1,2),(-3,4),(0,-4),(1,0)}

1. R is not a function because it is not single-valued. To element 1 in the souce corresponds 2 and 0 in the target set

2. Point (1,0) lies on y-axis

3. Point(-3,4) lies in quadrant II