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mathtard
 3 years ago
Can anyone please help me with this problem?
\[f(t)\sqrt{t^2+1}\]
find the function if it exists
mathtard
 3 years ago
Can anyone please help me with this problem? \[f(t)\sqrt{t^2+1}\] find the function if it exists

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estudier
 3 years ago
Best ResponseYou've already chosen the best response.2I suppose the function sqrt(t^2+1) does indeed exist...

mathtard
 3 years ago
Best ResponseYou've already chosen the best response.0but how do you find the function?

estudier
 3 years ago
Best ResponseYou've already chosen the best response.2U appear to have found it, it's right there...

estudier
 3 years ago
Best ResponseYou've already chosen the best response.2U are saying find the function but u are telling me what the function is?????

mathtard
 3 years ago
Best ResponseYou've already chosen the best response.0that is how the problem is written. ???????

estudier
 3 years ago
Best ResponseYou've already chosen the best response.2I dont understand what u want me to do, neither does anybody else.

mathtard
 3 years ago
Best ResponseYou've already chosen the best response.0well i am just writing the problem the way it is written.

mathtard
 3 years ago
Best ResponseYou've already chosen the best response.0oops find the function value if it exists

mathtard
 3 years ago
Best ResponseYou've already chosen the best response.0sorry i thought I wrote value .

estudier
 3 years ago
Best ResponseYou've already chosen the best response.2The only thing I can say about the function is that it must be positive by definition. Else you have to provide a t to enable calculation of a value for that particular value of t.

Alchemista
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(t)\sqrt{t^2+1} \cdot \sqrt{t^2+1}^{1} = f(t)\] which is the "value" of the function at t

mathtard
 3 years ago
Best ResponseYou've already chosen the best response.0sorry about that estudier.

Alchemista
 3 years ago
Best ResponseYou've already chosen the best response.0So you mean \[ f(t) = \sqrt{t^2+1}\]\[\text{Find $f(0)$}\]

Alchemista
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(0) = \sqrt{(0)^2+1} = \sqrt{1} = 1\]
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