Question

When solving (f of g)(4) do you solve for g first then f or vise versa? The given is f(x)=2x^2-1 and g(x) =6-1/2x^2
please and thank you (:

Answer 1

(f of g)(x) is f(g(x)), So you should start by solving g(x)=c and plugging that number into f(c). Understand?

Answer 2

ok let me try this

Answer 3

i got 8 for g of x, is that right?

Answer 4

*for (fog)(4)

Answer 5

so my final answer 127

Answer 6

let me know if that's right

Answer 7

so I got it right, correct?

Answer 8

cause i did get 31

Answer 9

oops nevermind i go that answer for (gof)(2)

Answer 10

hmmm... isn't g(4)=-2 ???

Answer 11

is this correct: \(\large g(x) = 6 - \frac{1}{2}x^2 \) ??

Answer 12

yes you are right i just realized i miscalculated

Answer 13

so now you need to calculate f(-2) ....

Answer 14

so the final answer must be 7

Answer 15

that's what i got....

Answer 16

yay!

Answer 17

good work... :)

Answer 18

wanna try another one and see of we get the same answer?

Answer 19

(fof)(1)

Answer 20

f(f(1)) = 1

Answer 21

I got 1 too

Answer 22

yay!!!

Answer 23

now (gof)(2)

Answer 24

i got 7 :)

Answer 25

(gog)(0)= 6 ?

Answer 26

g(f(2)) = g(7) = 6-(49/2) = -37/2 ...:(

Answer 27

what? i'm confused lol

Answer 28

for \(\large g(f(2)) = g(2\cdot \color {red}2^2-1)=g(7)=6-\frac{1}{2}\cdot 7^2=6-\frac{49}{2}=\frac{-37}{2} \)

Answer 29

\(\large g(g(0))=g(6-\frac{1}{2}\cdot \color {red}0^2)=g(6)=6-\frac{1}{2}\cdot 6^2=6-18=-9 \)

Answer 30

oops.. 6-18 = -12.... haha...

Answer 31

see I got stuck at 6-1/2*7^2 cause when i multiplied 49 times 1/2 i got 24.5

Answer 32

yes... that is correct... 49/2 = 24.5

Answer 33

so 6 - 24.5 = -18.5 = -37/2

Answer 34

so the answer could be either -18.5 or -37/2?

Answer 35

they have the same value.... they're equal...

it's like saying 3/2 = 1.5

Answer 36

oh ok i see :)

Answer 37

can you help me with 4 more problems or are you tired?

Answer 38

go ahead... post as a new question...

Answer 39

separately....

Answer 40

this ones kinda difficult, well for me at least

Answer 41

i had to take a screenshot of it

Answer 42

\(\large f(x)=x^2 \) and \(\large g(x)=x^2+1 \)

for a),
\(\large f(\color {red}{g(x)})=f(\color {red}{x^2+1})=(\color {red}{x^2+1})^2 \)
if you want, you can expand this...

Answer 43

this will give you a polynomial... the domain of all polynomials is all real numbers.

Answer 44

now you try part b, g(f(x)) = ???

Answer 45

g(f(x))=(x^2)^2+1 ?

Answer 46

yes... :)

Answer 47

so does this mean the same thing as the previous one, all real numbers?

Answer 48

yep... g(f(x)) is still a polynomial...

Answer 49

what makes it a polynomial?

Answer 50

the exponent in each term when expanded is 0 or a positive integer.

Answer 51

can you give me an example of a non polynomial equation

Answer 52

in your last problem, (x^2)^2 + 1 = x^4 + 1.... each term has a positive whole number exponent...

Answer 53

so if an equation has a negative or a fraction that makes it a non-polynomial?

Answer 54

ok....
non-polynomial: \(\large 45x^3-3x^2+x+7x^{-1} \)

Answer 55

if what you mean by fraction is like this: \(\large \frac{7}{x} \) , then yes, because

notice that \(\large 7x^{-1}=\frac{7}{x} \)

Answer 56

oh ok

Answer 57

http://www.mathsisfun.com/definitions/polynomial.html

Answer 58

thank you for the info

Answer 59

now we have (fof)(x) = (x^2)^2

Answer 60

yw...
what about f(f(x) =

Answer 61

oops.. you're way ahead of me... :)
yes.. that's correct...

Answer 62

im gonna have to bookmark that page you sent me it's very useful

Answer 63

lol

Answer 64

google does wonders!

Answer 65

so we don't have to worry about non-polynomials here huh since the given is pretty much a polynomial already

Answer 66

yes... anytime you have functions that are polynomials, their composition will always give you polynomials....
i think that's a theorem in algebra but not sure....

Answer 67

oh ok gotcha

Answer 68

ok last one for this problem is gog(x)
and i got (x^2+1)^2+1

Answer 69

yep... correct...
if u want, you can expand that....

Answer 70

how so

Answer 71

(x^2+1)(x^2+1)+1?

Answer 72

yes... FOIL it out and add the 1....

Answer 73

x^4+x^2+2

Answer 74

yep...

Answer 75

all polynomials will look like this:

\(\large a_nx^n+a_{n-1}x^{n-1}+...+a_2x^2+a_1x^1+a_0 \)

Answer 76

yes, i've seen that before, i was terrified when i first saw that lol

Answer 77

where the a's are just coefficients....

Answer 78

i'm still learning about the subs

Answer 79

that's where i kinda get stuck

Answer 80

it just means they could be (and usually is) different numbers in front of the x's....

Answer 81

okayy, ready for the next problem?

Answer 82

it's just math notation... don't let that worry you...

Answer 83

g'head...

Answer 84

\[x^3-x \le0\]

Answer 85

find the solution set?

Answer 86

it says solve for the inequality

Answer 87

yes

Answer 88

ok... first you need to solve the equation: \(\large x^3-x=0 \)
solve for x .... wha u got???

Answer 89

u want a hint?

Answer 90

\[x^3=x\] ?

Answer 91

not sure

Answer 92

no....

Answer 93

factor that left side completely....

Answer 94

\(\large x^3-x=x(x^2-1)=x(x+1)(x-1) \)

so the equation is: \(\large x(x+1)(x-1)=0 \)
now it's obvious what x is...

Answer 95

\[x(x-1)(x+1)\]

Answer 96

yes... what's x = ???

Answer 97

x is 0, 1 and -1?