Question

how do you find if: f(x)=2x-1, g(x)=x^2+1,
h(x)=5-3x...find (f+g)(x)

Answer 1

\[(f+g)(x)=f(x)+g(x)=2x-1+x^2+1=\cdots\]

Answer 2

how do i combine them again?

Answer 3

well the two constant terms can be simplified

Answer 4

yep i got that, the answer i got was \[2x+x^{2}\] is this right?

Answer 5

yep that is Right \(\checkmark\)

Answer 6

yay! okay, now what if its subtracting (f-g)(x)

Answer 7

\[(f-g)(x)=f(x)-g(x)=\cdots\]

Answer 8

\[f\circ g(x)=f(g(x))\]

Answer 9

hahah i get that, I mean i get confused with all the minus- so (2x-1)-(x^2+1)

Answer 10

your right so far,
now expand the brackets and remember to distribute the negative for the term on the right

Answer 11

2x-2-x^2.. thats what i got and I dont think thats right

Answer 12

that is right
\[(f-g)(x)=2x-2-x^2\qquad\checkmark\]

Answer 13

yes!! im actually understanding, okay Im having a little trouble with (f/g)(x)

Answer 14

\[ f(x)=2x-1;\qquad g(x)=x^2+1\]
\[(f/g)(x)=f(x)\div g(x)=\frac{2x-1}{x^2+1}=\cdots\]

Answer 15

okay i got that far, what do i do after i cancel out the -1 and the +1

Answer 16

the only way to simplify this is to break the fraction if to two components
like this
\[\frac{a+b}{c}=\frac ac+\frac bc\]

Answer 17

you can only cancel Factors common to the numerator and denominator in a fraction

Answer 18

okay so im really confused. I was going to guess 2x/x^2

Answer 19

\[(f/g)(x)=\frac{2x-1}{x^2+1}=\cdots\]
\[\frac{a+b}{c}=\frac ac+\frac bc\]
\[a=2x~;\qquad b=-1~;\qquad c=x^2+1\]

Answer 20

alright! I got that part written down, i just dont know what to do after that

Answer 21

make the substitution

Answer 22

can i cancel both +1's?

Answer 23

no, the 1's stay

Answer 24

I don't know

Answer 25

oh wait

Answer 26

can i do something with both denominators?

Answer 27

say if i had \[\frac {2y-3}{y^4+7}\] i could use the substitution\[\frac{a+b}{c}=\frac ac+\frac bc\]
to arrive at\[\frac {2y-3}{y^4+7}=\frac {2y}{y^4+7}-\frac3{y^4+7}\]
this cannot be simplified further

Answer 28

1x/2x^2+2... or am i completely wrong?

Answer 29

your answer should be two fractions

Answer 30

ohhh so its 2x/x^2+1 - -1/x^2+1

Answer 31

yes, but with only one negative sign

Answer 32

yeah thats just my subtract sign in the middle

Answer 33

if you mean this\[\frac{2x}{x^2+1} - \frac{1}{x^2+1}\]you are correct

Answer 34

oh I see what your saying!

Answer 35

will you check one for me to see if i did it right?

Answer 36

sure

Answer 37

(f of h) (2x-1)(5-3x)

Answer 38

is my answer -6x-5?

Answer 39

thats not right

Answer 40

-6x^2?

Answer 41

6x^2+7x-5?

Answer 42

\[f(x)=2x-1\qquad \qquad h(x)=5-3x\]
\[(f \circ h)=f(h(x))\]\[\qquad =f(5-3x)\]\[\qquad=2(5-3x)-1\]

Answer 43

When given the question: find f(g(x)), we plug in f(x) into g(x).true or false?

Answer 44

When given the question: find f(g(x)), we plug in f(x) into g(x).true or false?

Answer 45

shoot i got this mixed up with multiplication...

Answer 46

not quite,
we find g(x)
and then find f(g(x))

Answer 47

stop NOELL! im getting help

Answer 48

shes my sister...

Answer 49

the product of functions is\[f\times g(x)=f(x)\times g(x)\]
_________
\[f\circ g(x)=f(g(x))\]
is called the composition of functions

Answer 50

?

Answer 51

hello

Answer 52

noniebug is my sister. sorry! but okay back to me! would the answer be 6x-8?

Answer 53

that is \(h\circ f\)

Answer 54

\[h\circ f\neq f\circ h\]

Answer 55

8-6x

Answer 56

nope

Answer 57

uhhh im confused then.

Answer 58

\[(f \circ h)=f(h(x)) =f(5-3x) =2(5-3x)-1=\cdots\]

Answer 59

do i put the 2 with everything?

Answer 60

\[2(5−3x)−1=(2\times 5)+(2\times-3x)-1\]

Answer 61

10-6x-1? I dont know

Answer 62

thats good, now simplify the constants

Answer 63

9-6x

Answer 64

thats is correct \(\checkmark\)

Answer 65

yay! okay so i see where i got off at, i distubuted the 2 with the -2

Answer 66

-1**

Answer 67

now i have a multiplication one hah, i skipped it on acciendent. (f x g)(x)

Answer 68

you distributed the 2 across onto the 5 and the -3x

Answer 69

\[f\times g(x)=f(x)\times g(x)\]

Answer 70

yeah but i went wrong because i kept distributing all the way to the -1 and i wasn't suppose to!

Answer 71

ah i see

Answer 72

but okay! (2x-1)(x^2+1)

Answer 73

yep

Answer 74

i get confused when mulitpling is it 2x^3 or 3x^2

Answer 75

just the first step?

Answer 76

\[2x\times x^2=2x^3\]

Answer 77

so is the answer 2x^3+2x-x^2

Answer 78

there is one more term

Answer 79

2x^3+2x-1x^2?

Answer 80

\[(2x-1)(x^2+1)=2x\times x^2+2x-x^2-1\]

Answer 81

2x+2x-1? the x^2"s cancel out?

Answer 82

which would make it 4x^2-1 or am I wrong?

Answer 83

no remember \(2x×x^2=2x^3\)

Answer 84

2x^3+2x-x^2-1..I dont know after that :(

Answer 85

thats is as simplified as you can get.

Answer 86

oh so thats it??? oh i tried to cancel the 1's out when its multiplication...ha

Answer 87

that is it

Answer 88

yipee! okay do you have time for one more? its like the (f of g) one..

Answer 89

if your quick,

Answer 90

kay (g of h)

Answer 91

g=x^2+1 and h=5-3x

Answer 92

\[g\circ h(x)=g(h(x))=\]
substitute h in

Answer 93

is it 5x^2-3x^3+1?

Answer 94

not quite

\[g(5-3x)=(5-3x)^2+1 \]

Answer 95

..3x^2+26???????

Answer 96

not right
\[(a+b)^2=(a+b)(a+b)\]\[=a\times a+a\times b+b\times a+b\times b\]\[=a^2+2ab+b^2\]

Answer 97

9x^2+25?

Answer 98

9x^2+26