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sparrow2
 one year ago
transform y=f(x) to y=f(x1)
sparrow2
 one year ago
transform y=f(x) to y=f(x1)

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sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1i need prove that it goes right to 1 unit

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Say \((a,b)\) is a point on \(f(x)\) before the transformation, then we have \[f(a) = b\tag{1}\] Next, plugin \(x=\color{red}{a+1}\) into the transformed function \(f(x1)\), \[f(\color{red}{a+1}1) = f(a) = b \tag{2}\] That means \((a,b)\) has moved to \((a+1, b)\) under the transformation \(f(x1)\). After the transformation, the \(x\) value is \(a+1\) which is exactly \(1\) unit to the right of\(a\). We can conclude the transformation shifts \(x\) coordinates of all the points to right by \(1\) unit.

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1i understand that but here is what i don't undertand: an argument of the transformed function is x1, so when you plug for x=a+1, argument is a+11=a not a+1. @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Right, but my reference is still the "x" axis, not "x1" axis.

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1x axis mean that we put arguments there(not the meaning of x, like in y=f(2x) i put 2x meaning not x to x axis) right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Kindof, f(2x) compresses the x axis by a factor of 2, the reference is still x axis

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1i think it doesn't matter what we call x or not.. main is that we put arguments there.f(2z) i put the meaning of 2z to x axis too

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1in our case argument is a+11=a not a+1 so we have to put a not a+1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the argument to the transformed function is a+1

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1it is f(x1) you plug x=a+1(this is not the argument) argument is x1=a+11=a i see this like that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436472503364:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2argument is what ever goes into the function box from outside world output is whatever the function box spits out

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2btw it helps to treat f(x) and f(x1) as two different functions maybe just call g(x) = f(x1)

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436472582326:dw

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1i think here is main that f is the same (so the rule is the same)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436472673761:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Think of f(x) and f(x1) as two separate things

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1it is main that rules are the same( functions are different because they have different domains and range(mayby)

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1rules being the same is important

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I see what you're getting at, im still thinking of a better way to convince myself haha

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1this eats my brain :) just i think it is more theoritocal question

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Example: \[f(x)=x^2 \\ f(x1)=(x1)^2 \\ \text{ Let } g(x)=(x1)^2 \\ g(x+1)=(x+11)^2=x^2 =f(x) \\ \text{ the argument for } g \text{ was } x+1 \text{ on that last line }\] I'm not sure this answers your question or not. I'm still struggling with what you were talking about the other day. I'm not trying to be slow but what is it exactly you are looking clarification on?

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1i'm still confused..it will be good if you show the transformation from f(x) to f(x1) ,note that in f(x1) an argument is x1 not x,
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