Question

Could someone please check my answers?

Answer 1

f(x) = sqrt(7x-2) find the inverse function.
I got f^-1(x) = x^2+2/7

Answer 2

Find the domain f(x) = sqrt(2x-1/3x+4)
I got (-infinite,-4/3) U (1/2,infinite)

Answer 3

\(\large\color{slate}{ f(x)=\sqrt{7x-2} }\)
replace f(x) with y (since f(x) is same thing as y)
\(\large\color{slate}{ y=\sqrt{7x-2} }\)
replace x and y with each other.
\(\large\color{slate}{ x=\sqrt{7y-2} }\)
raise both sides to the second power
\(\large\color{slate}{ x^2=7y-2 }\)
add 2 to both sides,
\(\large\color{slate}{ x^2+2=7y }\)
divide both sides by 7,
\(\large\color{slate}{ \displaystyle \frac{x^2+2}{7}=y }\)
the notation for an inverse function is \(\normalsize\color{slate}{ \displaystyle f^{-1}(x)}\) so, write:
\(\large\color{slate}{ \displaystyle f^{-1}(x)=\frac{x^2+2}{7} }\)

YES, WELL DONE !

Answer 4

f(x) = 1/x^2-9 and g(x) = sqrt(x-2) f(g(x)) (x) find the domain
I got (-infinite, -11) U (-11, infinite)

Answer 5

\(\large\color{slate}{ \displaystyle f(x)=\sqrt{ \frac{2x-1}{3x+4} } }\) this is your function for number 2?

Answer 6

yes

Answer 7

find the domain and range of f^-1 given f(x) = 1/4-9x
I got Domain (-infinite,0) U (0,infinite) and Range (-infinite,4/9) U (4/9,infinite)

Answer 8

\(\large\color{slate}{ \displaystyle f(x)=\sqrt{ \frac{2x-1}{3x+4} } }\)


\(\large\color{slate}{ \displaystyle (- \infty, -4/3 ) ~~~{\rm U}~~~[1/2~,~\infty) }\)

is correct.
Because for any numbers less than -4/3 you get negative on top and bottom, which means that you have a positive inside the big root.

for numbers between -4/3 and 1/2 you will have only the top negative and bottom positive, which means negative in the square root (and that is permitted)
`at 1/2 you get y=0 (so it is just an x-intercept) don't see any reason not to include 1/2`.

and up to positive infinity from 1/2 you are getting positive over positive which is also fine.

Answer 9

f(x) = 1/x^2 and g(x) = sqrt(2+x) find the domain for f(g(x)) (x)
I got (-infinite,-2) U (-2,infinite)

Answer 10

@SolomonZelman you are helping me so much right now by checking my questions I am so grateful for you to take your time to help me.

Answer 11

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-9} }\)

\(\large\color{black}{ \displaystyle g(x) = \sqrt{x-2} }\)

and you want to find \(\large\color{black}{ \displaystyle g(f(x)) }\) and it's domain ?

Answer 12

Yes

Answer 13

Well just it's domain

Answer 14

\(\large\color{black}{ \displaystyle g(f(x))=\sqrt{\frac{1}{x^2-9}-2} }\)

this is the g(f(x)).
So, you need 2 things:

1) \(\large\color{black}{ \displaystyle x^2-9 \ne0 }\)

2) \(\large\color{black}{ \displaystyle \frac{1}{x^2-9}\ge2 }\)

Answer 15

so negative infinity makes the fraction in the root very small, leaving the square root negative.

Answer 16

and that is not allowed.

Answer 17

because with +inf or -inf, you have a large positive on the bottom, which makes everything in the root approach to (approx) -2.

Answer 18

oh, you said f(g(x)), my bad

Answer 19

(f o g) (x) is what they put

Answer 20

\(\large\color{black}{ \displaystyle f(g(x))=\frac{1}{(\sqrt{x-2})^2-9} }\)

\(\large\color{black}{ \displaystyle f(g(x))=\frac{1}{x-2-9} }\)

\(\large\color{black}{ \displaystyle f(g(x))=\frac{1}{x-11} }\)

Answer 21

but then, you are excluding not -11 from the domain, but positive 11

Answer 22

because at +11 the denominator is =0

Answer 23

oh okay

Answer 24

so it should be saying \(\large\color{black}{ \displaystyle (-\infty,11)~~{\rm U}~~ (-11,\infty) }\)

Answer 25

I put -11, let me redo it

Answer 26

\(\large\color{black}{ \displaystyle (-\infty,11)~~{\rm U}~~~(11,~\infty) }\)

Answer 27

like that

Answer 28

Evaluate the function f(x) = 2x - 5 and simplify at the indicated value f(-3) = ?
I got f(-3) = 1

Answer 29

thank you

Answer 30

for correcting it for me

Answer 31

```
find the domain and range of f^-1 given f(x) = 1/4-9x
I got Domain (-infinite,0) U (0,infinite) and Range (-infinite,4/9) U (4/9,infinite)
```

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{4-9x} }\)

\(\large\color{black}{ \displaystyle x=\frac{1}{4-9y} }\)

\(\large\color{black}{ \displaystyle 4-9y=\frac{1}{x} }\)

\(\large\color{black}{ \displaystyle -9y=\frac{1}{x}-\frac{4x}{x} }\)

\(\large\color{black}{ \displaystyle f^{-1}(x)=\frac{1-4x}{-9x} }\)

\(\large\color{black}{ \displaystyle f^{-1}(x)=\frac{4x-1}{9x} }\)

Answer 32

the domain and range you have are correct!

Answer 33

now what is next?

Answer 34

Yes!

Answer 35

```
f(x) = 1/x^2 and g(x) = sqrt(2+x) find the domain for f(g(x)) (x)
I got (-infinite,-2) U (-2,infinite)
```
this?

Answer 36

that is correct.

Answer 37

next?

Answer 38

yes

Answer 39

```
Evaluate the function f(x) = 2x - 5 and simplify at the indicated value f(-3) = ?
I got f(-3) = 1

Answer 40

yes

Answer 41

check this one please.

Answer 42

2 times -3 is +6 ?

Answer 43

no 2x

Answer 44

\(\large\color{slate}{ f(\color{red}{x}) =2\color{red}{x}-5 }\)
\(\large\color{slate}{ f(\color{red}{3}) =2\color{red}{(3)}-5 =6-5=1 }\)

f(3) is 1.

Answer 45

because f(3) gives `(POSITIVE 6 ) -5`.
but f(-3) gives `(NEGATIVE 6) -5`.

Answer 46

The function f(x) = -x^2 + 10 is not one to one. Find a portion of the domain where the function is one to one and finf an inverse function

Answer 47

either side of the parabola

Answer 48

from 0 to left or from 0 to right

Answer 49

yes