anonymous
  • anonymous
Range of\[f(x)=\sqrt[3]{x-1}+\sqrt[3]{3-x}\]
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
????
anonymous
  • anonymous
made a stupid error, just a sec.
anonymous
  • anonymous
|dw:1347772271638:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
yeah range is limited...i dont how to evaluate the exact interval for range :(
anonymous
  • anonymous
*know
anonymous
  • anonymous
Can you plot the two parts of the function? then combine them.|dw:1347772792991:dw| For analyzing the final result : If it's a diagonal line then yes both will go from -infinity to +infinity, IF HOWEVER it is a VERTICAL or HORIZONTAL line then your domain or range will be limited.
anonymous
  • anonymous
i want to show it algebrically if its possible !!
anonymous
  • anonymous
idea: seperate f(x) into 2 functions, find the domain of the inverse of each one of the function, find the intersection of the 2 domains, that gives the range of f(x)
nipunmalhotra93
  • nipunmalhotra93
Range is ]0,2]. f(x) is symmetrical about x=2. I evaluated the first and second derivatives. for x<1<2 f'(x)>0 and f''(x)>0 for x<1, f'(x)>0 and f''(x)<0. behavior on other side of 2 is symmetrical. Also, evaluating the limit at -infinity, we get 0. So the graph would be something like this.|dw:1347780606499:dw| That's it. Correct me if I'm wrong.
nipunmalhotra93
  • nipunmalhotra93
Another small test is that f(x) can't be <0 because solving it gives -1<-3 which is a contradiction.
anonymous
  • anonymous
the range of f(x) is ( 0,2]
anonymous
  • anonymous
to show it algebrically you need to assume \[\sqrt[3]{x-1}= \alpha\] \[\sqrt[3]{3-x}= \beta\] and after that you can obtain the required equations
anonymous
  • anonymous
as \[f(x)=y=\alpha +\beta\] so find out x in form of y and you would get the answer if you want more detailed solution then i can provide you that too

Looking for something else?

Not the answer you are looking for? Search for more explanations.