Question

calculate g(x), where g(x) is the inverse of f(x).
f(x)=x−3 calculate g(x), where g(x) is the inverse of f(x).
f(x)=x−3 @Mathematics

Answer 1

f(g(x)) = x then

Answer 2

f(x)=x−3

y=x−3


x=y−3


x+3 = y


y = x+3


So the inverse is g(x) = x+3

Answer 3

g(x) - 3 = x when g(x) = x+3

Answer 4

@amistre can u help me after ur done
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Answer 5

I thought it was too, but when I enter it as answer it says its wrong

Answer 6

well there must be a typo somewhere

Answer 7

it asks me to use theorem 1 from the text to solve it

Answer 8

and what is theorem 1 in the text?

Answer 9

g'(b)= 1/f'(g(b))

Answer 10

and what does the " ' " ticl mean here?

Answer 11

g prime and f prime

Answer 12

derivative of.

Answer 13

i dont understand how that plays into the question really; id have to read your material

Answer 14

yeah me neither, hence my confusion.

Answer 15

Use Theorem 1 from the text to calculate g(x), where g(x) is the inverse of f(x).
f(x)=x−3

Answer 16

g'(b)=(1)/f'(g(b))

Answer 17

[f(g(x))]' = f'(g(x)) g'(x)

f'(x) = 1; g(x) = 1 so g'(x) = 0 ....

1/(f'(1)) ..... jibberjabber is all i see

Answer 18

\[\frac{dy}{dx}=\frac{1}{dx/dy}\] is a thrm

Answer 19

f'(g(x)) g'(x) = 1 from the chain rule

g'(x) = 1/f'(g(x))

Answer 20

is where its from

Answer 21

\[g'(b)=1\div f'(g(b))\]

Answer 22

f(x) = x-3 ; g(x) = x+3
f'(x) = 1 ; g'(x) = 1

g'(x) = 1/f'(g(x))

1 = 1/f'(x+3)

f'(x+3) = 1

its gotta be something to do with these

Answer 23

joemath might remember it

Answer 24

yeah the text really confused me.