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if v= (ax^2 + b)^ 0.5 would the variation of v with x be a x^2 type variation or a root X type variation?? i feel that it should depend on the value of constant a since d over dx of dv/dx can be negative or positive depending on value of a.. however the book says that it shuld be a x^2 type variation.. kindly help

if v= (ax^2 + b)^ 0.5


 


would the variation of v with x be a x^2 type variation or a root X type variation??


i feel that it should depend on the value of constant a since d over dx of dv/dx can be negative or positive depending on value of a..


 


however the book says that it shuld be a x^2 type variation..


 


kindly help

Grade:

1 Answers

Prudhvi teja
83 Points
11 years ago

Dear parth

a and b cant be negative

because it is defined for all x so

when x=0

v should exist so b>0

if a is negative at infinity v doesnt exist

so both a and b should exist.

and when look at graph you will get that it will have x2 variation

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Prudhvi Teja

 

 


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