I got a small problem Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Maximum of functions word problemProblem in understanding composite functionsHow do I solve this function and find its domain?Composition of functions, examplesProblem with composite relationsBehaviour of composition functions of a composite functionA specific problem about composition of functionsFunction composition and bijectionMapping problem involving equivalence relationsWhat does it mean for a composite function to be defined?
Using audio cues to encourage good posture
Why aren't air breathing engines used as small first stages?
Trademark violation for app?
Selecting user stories during sprint planning
How to install press fit bottom bracket into new frame
Is CEO the "profession" with the most psychopaths?
How do I use the new nonlinear finite element in Mathematica 12 for this equation?
What's the meaning of "fortified infraction restraint"?
What is the appropriate index architecture when forced to implement IsDeleted (soft deletes)?
How to tell that you are a giant?
What is the difference between globalisation and imperialism?
AppleTVs create a chatty alternate WiFi network
How to react to hostile behavior from a senior developer?
How do living politicians protect their readily obtainable signatures from misuse?
Generate an RGB colour grid
Why is my ESD wriststrap failing with nitrile gloves on?
Disembodied hand growing fangs
If Windows 7 doesn't support WSL, then what does Linux subsystem option mean?
Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?
Amount of permutations on an NxNxN Rubik's Cube
Why is it faster to reheat something than it is to cook it?
What do you call the main part of a joke?
What is a fractional matching?
Why should I vote and accept answers?
I got a small problem
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Maximum of functions word problemProblem in understanding composite functionsHow do I solve this function and find its domain?Composition of functions, examplesProblem with composite relationsBehaviour of composition functions of a composite functionA specific problem about composition of functionsFunction composition and bijectionMapping problem involving equivalence relationsWhat does it mean for a composite function to be defined?
$begingroup$
Let $f(X) = 5X + 4$, $g(X) = 4X + 3$. Suppose that $f circ g (X) = aX + b$. Find $a + b$. How can I find sum of $a$ and $b$
functions
edited Apr 1 at 19:04
Ernie060
2,940719
asked Apr 1 at 18:16
brandonbrandon
32
$endgroup$
add a comment |
$begingroup$
Let $f(X) = 5X + 4$, $g(X) = 4X + 3$. Suppose that $f circ g (X) = aX + b$. Find $a + b$. How can I find sum of $a$ and $b$
functions
edited Apr 1 at 19:04
Ernie060
2,940719
asked Apr 1 at 18:16
brandonbrandon
32
$endgroup$
$begingroup$
Have you tried computing $fcirc g$? $fcirc g(x) = f(g(x))=f(4x+3)=5(4x+3)+4$..
$endgroup$
– Don Thousand
Apr 1 at 18:17
$begingroup$
yes, i got 20X + 19 = aX+ b but i stuck right heree
$endgroup$
– brandon
Apr 1 at 18:18
$begingroup$
That means $a=20,b=19$... you did the hard part already.
$endgroup$
– Don Thousand
Apr 1 at 18:18
1
$begingroup$
oh god, that's so simple. Why can i not think about it, stupid me
$endgroup$
– brandon
Apr 1 at 18:19
$begingroup$
Well,.... there is the subtlety as to why that is an answer or the only answer... You can get $(20-a)X = 19-b$ and this must be true for all $X$s. So it it must be true for $X=0$. And it must be true for $X = 1$. So $(20-a)cdot 0 = 19-b$. There is only one answer for $b$. And $(20-a)cdot 1 = 19-b$. For that one answer for $b$ there is only one answer for $a$. ... But that's probably more details then you need.
$endgroup$
– fleablood
Apr 1 at 18:43
add a comment |
$begingroup$
Let $f(X) = 5X + 4$, $g(X) = 4X + 3$. Suppose that $f circ g (X) = aX + b$. Find $a + b$. How can I find sum of $a$ and $b$
functions
edited Apr 1 at 19:04
Ernie060
2,940719
asked Apr 1 at 18:16
brandonbrandon
32
$endgroup$
Let $f(X) = 5X + 4$, $g(X) = 4X + 3$. Suppose that $f circ g (X) = aX + b$. Find $a + b$. How can I find sum of $a$ and $b$
functions
functions
edited Apr 1 at 19:04
Ernie060
2,940719
asked Apr 1 at 18:16
brandonbrandon
32
edited Apr 1 at 19:04
Ernie060
2,940719
asked Apr 1 at 18:16
brandonbrandon
32
edited Apr 1 at 19:04
Ernie060
2,940719
edited Apr 1 at 19:04
Ernie060
2,940719
edited Apr 1 at 19:04
Ernie060
2,940719
2,940719
asked Apr 1 at 18:16
brandonbrandon
32
asked Apr 1 at 18:16
brandonbrandon
32
asked Apr 1 at 18:16
brandonbrandon
32
32
$begingroup$
Have you tried computing $fcirc g$? $fcirc g(x) = f(g(x))=f(4x+3)=5(4x+3)+4$..
$endgroup$
– Don Thousand
Apr 1 at 18:17
$begingroup$
yes, i got 20X + 19 = aX+ b but i stuck right heree
$endgroup$
– brandon
Apr 1 at 18:18
$begingroup$
That means $a=20,b=19$... you did the hard part already.
$endgroup$
– Don Thousand
Apr 1 at 18:18
1
$begingroup$
oh god, that's so simple. Why can i not think about it, stupid me
$endgroup$
– brandon
Apr 1 at 18:19
$begingroup$
Well,.... there is the subtlety as to why that is an answer or the only answer... You can get $(20-a)X = 19-b$ and this must be true for all $X$s. So it it must be true for $X=0$. And it must be true for $X = 1$. So $(20-a)cdot 0 = 19-b$. There is only one answer for $b$. And $(20-a)cdot 1 = 19-b$. For that one answer for $b$ there is only one answer for $a$. ... But that's probably more details then you need.
$endgroup$
– fleablood
Apr 1 at 18:43
add a comment |
$begingroup$
Have you tried computing $fcirc g$? $fcirc g(x) = f(g(x))=f(4x+3)=5(4x+3)+4$..
$endgroup$
– Don Thousand
Apr 1 at 18:17
$begingroup$
yes, i got 20X + 19 = aX+ b but i stuck right heree
$endgroup$
– brandon
Apr 1 at 18:18
$begingroup$
That means $a=20,b=19$... you did the hard part already.
$endgroup$
– Don Thousand
Apr 1 at 18:18
1
$begingroup$
oh god, that's so simple. Why can i not think about it, stupid me
$endgroup$
– brandon
Apr 1 at 18:19
$begingroup$
Well,.... there is the subtlety as to why that is an answer or the only answer... You can get $(20-a)X = 19-b$ and this must be true for all $X$s. So it it must be true for $X=0$. And it must be true for $X = 1$. So $(20-a)cdot 0 = 19-b$. There is only one answer for $b$. And $(20-a)cdot 1 = 19-b$. For that one answer for $b$ there is only one answer for $a$. ... But that's probably more details then you need.
$endgroup$
– fleablood
Apr 1 at 18:43
$begingroup$
Have you tried computing $fcirc g$? $fcirc g(x) = f(g(x))=f(4x+3)=5(4x+3)+4$..
$endgroup$
– Don Thousand
Apr 1 at 18:17
$begingroup$
Have you tried computing $fcirc g$? $fcirc g(x) = f(g(x))=f(4x+3)=5(4x+3)+4$..
$endgroup$
– Don Thousand
Apr 1 at 18:17
$begingroup$
yes, i got 20X + 19 = aX+ b but i stuck right heree
$endgroup$
– brandon
Apr 1 at 18:18
$begingroup$
yes, i got 20X + 19 = aX+ b but i stuck right heree
$endgroup$
– brandon
Apr 1 at 18:18
$begingroup$
That means $a=20,b=19$... you did the hard part already.
$endgroup$
– Don Thousand
Apr 1 at 18:18
$begingroup$
That means $a=20,b=19$... you did the hard part already.
$endgroup$
– Don Thousand
Apr 1 at 18:18
1
1
$begingroup$
oh god, that's so simple. Why can i not think about it, stupid me
$endgroup$
– brandon
Apr 1 at 18:19
$begingroup$
oh god, that's so simple. Why can i not think about it, stupid me
$endgroup$
– brandon
Apr 1 at 18:19
$begingroup$
Well,.... there is the subtlety as to why that is an answer or the only answer... You can get $(20-a)X = 19-b$ and this must be true for all $X$s. So it it must be true for $X=0$. And it must be true for $X = 1$. So $(20-a)cdot 0 = 19-b$. There is only one answer for $b$. And $(20-a)cdot 1 = 19-b$. For that one answer for $b$ there is only one answer for $a$. ... But that's probably more details then you need.
$endgroup$
– fleablood
Apr 1 at 18:43
$begingroup$
Well,.... there is the subtlety as to why that is an answer or the only answer... You can get $(20-a)X = 19-b$ and this must be true for all $X$s. So it it must be true for $X=0$. And it must be true for $X = 1$. So $(20-a)cdot 0 = 19-b$. There is only one answer for $b$. And $(20-a)cdot 1 = 19-b$. For that one answer for $b$ there is only one answer for $a$. ... But that's probably more details then you need.
$endgroup$
– fleablood
Apr 1 at 18:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$fcirc g(x) = f(g(x))$.
$f(whatever) = 5(whatever) + 4$ so $f(g(x)) = 5(g(x)) + 4$.
And $g(x) = 4x + 3$. So
$f(g(x)) = 5(g(x)) + 4 = 5(4x + 3) + 4$.
We are told $fcirc g(x) = ax + b$.
So $fcirc g(x) = 5(4x+3) + 4 = ax + b$.
So what is $a$ and $b$?
It's a straight forward question.
answered Apr 1 at 18:37
fleabloodfleablood
1
$endgroup$
add a comment |
Your Answer
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%2f3170943%2fi-got-a-small-problem%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
$begingroup$
$fcirc g(x) = f(g(x))$.
$f(whatever) = 5(whatever) + 4$ so $f(g(x)) = 5(g(x)) + 4$.
And $g(x) = 4x + 3$. So
$f(g(x)) = 5(g(x)) + 4 = 5(4x + 3) + 4$.
We are told $fcirc g(x) = ax + b$.
So $fcirc g(x) = 5(4x+3) + 4 = ax + b$.
So what is $a$ and $b$?
It's a straight forward question.
answered Apr 1 at 18:37
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
$fcirc g(x) = f(g(x))$.
$f(whatever) = 5(whatever) + 4$ so $f(g(x)) = 5(g(x)) + 4$.
And $g(x) = 4x + 3$. So
$f(g(x)) = 5(g(x)) + 4 = 5(4x + 3) + 4$.
We are told $fcirc g(x) = ax + b$.
So $fcirc g(x) = 5(4x+3) + 4 = ax + b$.
So what is $a$ and $b$?
It's a straight forward question.
answered Apr 1 at 18:37
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
$fcirc g(x) = f(g(x))$.
$f(whatever) = 5(whatever) + 4$ so $f(g(x)) = 5(g(x)) + 4$.
And $g(x) = 4x + 3$. So
$f(g(x)) = 5(g(x)) + 4 = 5(4x + 3) + 4$.
We are told $fcirc g(x) = ax + b$.
So $fcirc g(x) = 5(4x+3) + 4 = ax + b$.
So what is $a$ and $b$?
It's a straight forward question.
answered Apr 1 at 18:37
fleabloodfleablood
1
$endgroup$
$fcirc g(x) = f(g(x))$.
$f(whatever) = 5(whatever) + 4$ so $f(g(x)) = 5(g(x)) + 4$.
And $g(x) = 4x + 3$. So
$f(g(x)) = 5(g(x)) + 4 = 5(4x + 3) + 4$.
We are told $fcirc g(x) = ax + b$.
So $fcirc g(x) = 5(4x+3) + 4 = ax + b$.
So what is $a$ and $b$?
It's a straight forward question.
answered Apr 1 at 18:37
fleabloodfleablood
1
answered Apr 1 at 18:37
fleabloodfleablood
1
answered Apr 1 at 18:37
fleabloodfleablood
1
answered Apr 1 at 18:37
fleabloodfleablood
1
1
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%2f3170943%2fi-got-a-small-problem%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
$begingroup$
Have you tried computing $fcirc g$? $fcirc g(x) = f(g(x))=f(4x+3)=5(4x+3)+4$..
$endgroup$
– Don Thousand
Apr 1 at 18:17
$begingroup$
yes, i got 20X + 19 = aX+ b but i stuck right heree
$endgroup$
– brandon
Apr 1 at 18:18
$begingroup$
That means $a=20,b=19$... you did the hard part already.
$endgroup$
– Don Thousand
Apr 1 at 18:18
1
$begingroup$
oh god, that's so simple. Why can i not think about it, stupid me
$endgroup$
– brandon
Apr 1 at 18:19
$begingroup$
Well,.... there is the subtlety as to why that is an answer or the only answer... You can get $(20-a)X = 19-b$ and this must be true for all $X$s. So it it must be true for $X=0$. And it must be true for $X = 1$. So $(20-a)cdot 0 = 19-b$. There is only one answer for $b$. And $(20-a)cdot 1 = 19-b$. For that one answer for $b$ there is only one answer for $a$. ... But that's probably more details then you need.
$endgroup$
– fleablood
Apr 1 at 18:43