If $(fcirc g)(x)=tan^2x$ and $g(x)=sqrt{cos2x}$ then find $f(x)=?$
We've given :
$$(fcirc g)(x)=tan^2x$$
and
$$g(x)=sqrt{cos 2x}$$
Then how to find the function $f(x)$?
I know that
$$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
But I do not know how to find $f(x)$!
Please help me!
algebra-precalculus functions function-and-relation-composition
edited Dec 8 at 18:23
John Hughes
62.3k24090
asked Dec 8 at 18:12
user602338
1587
add a comment |
We've given :
$$(fcirc g)(x)=tan^2x$$
and
$$g(x)=sqrt{cos 2x}$$
Then how to find the function $f(x)$?
I know that
$$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
But I do not know how to find $f(x)$!
Please help me!
algebra-precalculus functions function-and-relation-composition
edited Dec 8 at 18:23
John Hughes
62.3k24090
asked Dec 8 at 18:12
user602338
1587
add a comment |
We've given :
$$(fcirc g)(x)=tan^2x$$
and
$$g(x)=sqrt{cos 2x}$$
Then how to find the function $f(x)$?
I know that
$$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
But I do not know how to find $f(x)$!
Please help me!
algebra-precalculus functions function-and-relation-composition
edited Dec 8 at 18:23
John Hughes
62.3k24090
asked Dec 8 at 18:12
user602338
1587
We've given :
$$(fcirc g)(x)=tan^2x$$
and
$$g(x)=sqrt{cos 2x}$$
Then how to find the function $f(x)$?
I know that
$$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
But I do not know how to find $f(x)$!
Please help me!
algebra-precalculus functions function-and-relation-composition
algebra-precalculus functions function-and-relation-composition
edited Dec 8 at 18:23
John Hughes
62.3k24090
asked Dec 8 at 18:12
user602338
1587
edited Dec 8 at 18:23
John Hughes
62.3k24090
asked Dec 8 at 18:12
user602338
1587
edited Dec 8 at 18:23
John Hughes
62.3k24090
edited Dec 8 at 18:23
John Hughes
62.3k24090
edited Dec 8 at 18:23
John Hughes
62.3k24090
62.3k24090
asked Dec 8 at 18:12
user602338
1587
asked Dec 8 at 18:12
user602338
1587
asked Dec 8 at 18:12
user602338
1587
1587
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
So you can find
$$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$
and
$$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$
obviously
$$f(x)=frac{1-x^2}{1+x^2}$$
answered Dec 8 at 18:22
Nanayajitzuki
1885
add a comment |
Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".
Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$
Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
add a comment |
You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$
answered Dec 8 at 18:18
Andrei
11k21025
I don't see how this gives $f(x)$.
– Shaun
Dec 8 at 18:23
$f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
– Andrei
Dec 8 at 18:25
Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
– Shaun
Dec 8 at 18:28
Just look at the other answers. They explicitly follow the steps that I've described
– Andrei
Dec 8 at 18:30
Ah, okay; thanks again :)
– Shaun
Dec 8 at 18:31
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%2f3031429%2fif-f-circ-gx-tan2x-and-gx-sqrt-cos2x-then-find-fx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
So you can find
$$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$
and
$$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$
obviously
$$f(x)=frac{1-x^2}{1+x^2}$$
answered Dec 8 at 18:22
Nanayajitzuki
1885
add a comment |
So you can find
$$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$
and
$$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$
obviously
$$f(x)=frac{1-x^2}{1+x^2}$$
answered Dec 8 at 18:22
Nanayajitzuki
1885
add a comment |
So you can find
$$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$
and
$$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$
obviously
$$f(x)=frac{1-x^2}{1+x^2}$$
answered Dec 8 at 18:22
Nanayajitzuki
1885
So you can find
$$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$
and
$$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$
obviously
$$f(x)=frac{1-x^2}{1+x^2}$$
answered Dec 8 at 18:22
Nanayajitzuki
1885
answered Dec 8 at 18:22
Nanayajitzuki
1885
answered Dec 8 at 18:22
Nanayajitzuki
1885
answered Dec 8 at 18:22
Nanayajitzuki
1885
1885
add a comment |
add a comment |
Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".
Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$
Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
add a comment |
Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".
Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$
Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
add a comment |
Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".
Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$
Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".
Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$
Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
answered Dec 8 at 18:21
Ivo Terek
45.2k951139
45.2k951139
add a comment |
add a comment |
You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$
answered Dec 8 at 18:18
Andrei
11k21025
I don't see how this gives $f(x)$.
– Shaun
Dec 8 at 18:23
$f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
– Andrei
Dec 8 at 18:25
Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
– Shaun
Dec 8 at 18:28
Just look at the other answers. They explicitly follow the steps that I've described
– Andrei
Dec 8 at 18:30
Ah, okay; thanks again :)
– Shaun
Dec 8 at 18:31
add a comment |
You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$
answered Dec 8 at 18:18
Andrei
11k21025
I don't see how this gives $f(x)$.
– Shaun
Dec 8 at 18:23
$f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
– Andrei
Dec 8 at 18:25
Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
– Shaun
Dec 8 at 18:28
Just look at the other answers. They explicitly follow the steps that I've described
– Andrei
Dec 8 at 18:30
Ah, okay; thanks again :)
– Shaun
Dec 8 at 18:31
add a comment |
You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$
answered Dec 8 at 18:18
Andrei
11k21025
You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$
answered Dec 8 at 18:18
Andrei
11k21025
answered Dec 8 at 18:18
Andrei
11k21025
answered Dec 8 at 18:18
Andrei
11k21025
answered Dec 8 at 18:18
Andrei
11k21025
11k21025
I don't see how this gives $f(x)$.
– Shaun
Dec 8 at 18:23
$f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
– Andrei
Dec 8 at 18:25
Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
– Shaun
Dec 8 at 18:28
Just look at the other answers. They explicitly follow the steps that I've described
– Andrei
Dec 8 at 18:30
Ah, okay; thanks again :)
– Shaun
Dec 8 at 18:31
add a comment |
I don't see how this gives $f(x)$.
– Shaun
Dec 8 at 18:23
$f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
– Andrei
Dec 8 at 18:25
Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
– Shaun
Dec 8 at 18:28
Just look at the other answers. They explicitly follow the steps that I've described
– Andrei
Dec 8 at 18:30
Ah, okay; thanks again :)
– Shaun
Dec 8 at 18:31
I don't see how this gives $f(x)$.
– Shaun
Dec 8 at 18:23
I don't see how this gives $f(x)$.
– Shaun
Dec 8 at 18:23
$f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
– Andrei
Dec 8 at 18:25
$f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
– Andrei
Dec 8 at 18:25
Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
– Shaun
Dec 8 at 18:28
Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
– Shaun
Dec 8 at 18:28
Just look at the other answers. They explicitly follow the steps that I've described
– Andrei
Dec 8 at 18:30
Just look at the other answers. They explicitly follow the steps that I've described
– Andrei
Dec 8 at 18:30
Ah, okay; thanks again :)
– Shaun
Dec 8 at 18:31
Ah, okay; thanks again :)
– Shaun
Dec 8 at 18:31
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3031429%2fif-f-circ-gx-tan2x-and-gx-sqrt-cos2x-then-find-fx%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
