Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Range of\[f(x)=\sqrt[3]{x-1}+\sqrt[3]{3-x}\]

I got my questions answered at in under 10 minutes. Go to now for free help!
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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

made a stupid error, just a sec.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yeah range is limited...i dont how to evaluate the exact interval for range :(
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.
i want to show it algebrically if its possible !!
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)
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.
Another small test is that f(x) can't be <0 because solving it gives -1<-3 which is a contradiction.
the range of f(x) is ( 0,2]
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
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question