Transformations on function curve
$begingroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
34k22455
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
$endgroup$
add a comment |
$begingroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
34k22455
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
$endgroup$
add a comment |
$begingroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
34k22455
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
$endgroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
34k22455
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
edited Dec 23 '18 at 23:49
gt6989b
34k22455
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
edited Dec 23 '18 at 23:49
gt6989b
34k22455
edited Dec 23 '18 at 23:49
gt6989b
34k22455
edited Dec 23 '18 at 23:49
gt6989b
34k22455
34k22455
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
335
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050784%2ftransformations-on-function-curve%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
$endgroup$
add a comment |
$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
$endgroup$
add a comment |
$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
$endgroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
answered Dec 23 '18 at 23:49
gt6989bgt6989b
34k22455
34k22455
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050784%2ftransformations-on-function-curve%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
