I know it a lot I’m just confused

Question

1)Find the domain of the function (f/x)(x)=x^2-4x-5 and g(x)=25

a){x|x≠-1or5}

b){x|x≠-5or5

c){x|x≠-1}

d){x|x≠-5}

2)find the (fog)x when f(x)=x^2+6+5 and g(x)=1/x+1

3)which function has asymptote at x=39

a)f(x)=7^x-39

b)f(x)=log7(x-39)

C)f(x)=7x^+39

d)f(x)=log7(x+39)

4)define a function that transforms the parent root function with a horizontal compression by a factor of 3 and a down ward shift of 15 units

5)what is the domian of f/g , given f(x) and g(x)=x-7

6)simplfy f+g/f-g when f(x)x-2/x+5 and g(x)x-5/x-2

7)which equation has exactly two real and two no real solutions

a)x^4-36x^2=0

b)x^3-x^2+x-4=0

c)x^4-5x^2-36=0

d)x^3-2x^2+4=0

8)determine whether each sequence below is arithmetic or not

sequence 1- 1/2,7/6,11/6,5/2

sequence2- 1/2,1/3,2/9,4/27

9)caluclate s20 for arithmetic sequence in which a11=3.2 and common difference is d=2.9

Answer 1

Hi Christina,

Here are some things to get you started.

1)There is some information that has typos, so it is not clear. Is it the case that 1/f(x) = x²-4x - 5? The first type to find the domain is to factor the expression and see for which values of x f(x) will not be defined.

x²-4x - 5 = x²-5x + x - 5 = x(x-5)+ (x-5) = (x-5) (x+1)

If you have 1/f(x) = x²-4x - 5, then f(x) = 1/ ((x-5) (x+1)). You cannot have a denominator that is 0, and thus x cannot be 5 or -1. Keep in mind that this holds only if the assumption about what the actual formula of f(x) is is correct.

2) (f°g) (x) = f(g(x)) = ?

Is it that f(x) = x²+6x +5? If so, you can factor it as: f(x) = x²+5x + x+5 = (x+5)(x²+1)

g(x) = 1/x + 1. Or is it g(x) = 1/(x+1)?

Thus f(g(x)) = f(1/x + 1) = (1/x + 1 + 5) * ((1/x + 1)^2+1) = (1/x + 6) * (1/x² + 2/x + 1+ 1) = (1/x + 6) * (1/x² + 2/x + 2)

You can unpack it from here.

Please consider working with a tutor. Am here if you have more questions. Good luck!