Find range of exponent

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












I have the following function I need to find the range for and I'm not sure if I'm on the right direction.



$f(x,y) = e^-x^2-(y-1)^2$



$x$ & $y$ are real-numbers.



I'm thinking that the range is "all real values for $y$ that are $> 0$."



Is this right?






algebra-precalculus functions exponential-function







share|cite|improve this question























  • I think you need to find the range of f(x,y), not y. You need to think about what all values f(x,y) can take for any real x, y
    – Neo
    39 mins ago














up vote
2
down vote

favorite












I have the following function I need to find the range for and I'm not sure if I'm on the right direction.



$f(x,y) = e^-x^2-(y-1)^2$



$x$ & $y$ are real-numbers.



I'm thinking that the range is "all real values for $y$ that are $> 0$."



Is this right?






algebra-precalculus functions exponential-function







share|cite|improve this question























  • I think you need to find the range of f(x,y), not y. You need to think about what all values f(x,y) can take for any real x, y
    – Neo
    39 mins ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have the following function I need to find the range for and I'm not sure if I'm on the right direction.



$f(x,y) = e^-x^2-(y-1)^2$



$x$ & $y$ are real-numbers.



I'm thinking that the range is "all real values for $y$ that are $> 0$."



Is this right?






algebra-precalculus functions exponential-function







share|cite|improve this question















I have the following function I need to find the range for and I'm not sure if I'm on the right direction.



$f(x,y) = e^-x^2-(y-1)^2$



$x$ & $y$ are real-numbers.



I'm thinking that the range is "all real values for $y$ that are $> 0$."



Is this right?






algebra-precalculus functions exponential-function




algebra-precalculus functions exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 22 mins ago









MRobinson

1,374218




1,374218










asked 43 mins ago









Hews

776




776











  • I think you need to find the range of f(x,y), not y. You need to think about what all values f(x,y) can take for any real x, y
    – Neo
    39 mins ago
















  • I think you need to find the range of f(x,y), not y. You need to think about what all values f(x,y) can take for any real x, y
    – Neo
    39 mins ago















I think you need to find the range of f(x,y), not y. You need to think about what all values f(x,y) can take for any real x, y
– Neo
39 mins ago




I think you need to find the range of f(x,y), not y. You need to think about what all values f(x,y) can take for any real x, y
– Neo
39 mins ago










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










Since $-x^2-(y-1)^2leq0$ and $g(x)=e^x$ increases, we obtain:
$$0<e^-x^2-(y-1)^2leq e^0=1.$$






share|cite|improve this answer






















  • Why does it need to be less than 1 though? - Thanks for the edit!
    – Hews
    37 mins ago







  • 1




    @Hews I added something. See now.
    – Michael Rozenberg
    34 mins ago










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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2951134%2ffind-range-of-exponent%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Since $-x^2-(y-1)^2leq0$ and $g(x)=e^x$ increases, we obtain:
$$0<e^-x^2-(y-1)^2leq e^0=1.$$






share|cite|improve this answer






















  • Why does it need to be less than 1 though? - Thanks for the edit!
    – Hews
    37 mins ago







  • 1




    @Hews I added something. See now.
    – Michael Rozenberg
    34 mins ago














up vote
3
down vote



accepted










Since $-x^2-(y-1)^2leq0$ and $g(x)=e^x$ increases, we obtain:
$$0<e^-x^2-(y-1)^2leq e^0=1.$$






share|cite|improve this answer






















  • Why does it need to be less than 1 though? - Thanks for the edit!
    – Hews
    37 mins ago







  • 1




    @Hews I added something. See now.
    – Michael Rozenberg
    34 mins ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Since $-x^2-(y-1)^2leq0$ and $g(x)=e^x$ increases, we obtain:
$$0<e^-x^2-(y-1)^2leq e^0=1.$$






share|cite|improve this answer














Since $-x^2-(y-1)^2leq0$ and $g(x)=e^x$ increases, we obtain:
$$0<e^-x^2-(y-1)^2leq e^0=1.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 35 mins ago

























answered 41 mins ago









Michael Rozenberg

90.8k1584181




90.8k1584181











  • Why does it need to be less than 1 though? - Thanks for the edit!
    – Hews
    37 mins ago







  • 1




    @Hews I added something. See now.
    – Michael Rozenberg
    34 mins ago
















  • Why does it need to be less than 1 though? - Thanks for the edit!
    – Hews
    37 mins ago







  • 1




    @Hews I added something. See now.
    – Michael Rozenberg
    34 mins ago















Why does it need to be less than 1 though? - Thanks for the edit!
– Hews
37 mins ago





Why does it need to be less than 1 though? - Thanks for the edit!
– Hews
37 mins ago





1




1




@Hews I added something. See now.
– Michael Rozenberg
34 mins ago




@Hews I added something. See now.
– Michael Rozenberg
34 mins ago

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2951134%2ffind-range-of-exponent%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What does second last employer means? [closed]

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
Read more

List of Gilmore Girls characters

Image
The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
Read more

Confectionery

Image
Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of ...
Read more