HELP PLEASE!! F(x)=x+(50/x^3) a)What is the domain of F(x)? Chose one answer 1. x does not equal to o and +-^4squareroot of 50 2. x does not equal 0 and ^4squareroot of 50 3. x does not equal 0 4. All real numbers b)What is the equation of the vertical asymptote(s) of F(x)? 1. x=__________ (use comma to separate answer as needed. Type an integer or a fraction) OR 2. there is no vertical asymptote c)What is the equation of the horizontal of oblique asymptote of F(x)? 1. y=________ OR 2. there is no horizontal or oblique
domain- find a value of x when f(x) has either 1. a division by 0 or 2. square-root of -ve number Domain is then all number except for those you just found.
confused still?
\[f(x)=x+{50\over x^3}\] when can you get a division by 0 here?
no
what if x=0?
but you can't divide by 0
exactly... so, you cannot have x=0 in the domain of f(x)
domain is any 'x' for which you can calculate f(x)
okay what about +- \[\sqrt[4]{50}\]
it is still a number. 1. does it give you a division by 0? 2. does it give you a sqre-root of -ve number? NO so, \[\sqrt[4]{50}\] is a valid number.
ooooh! okay
but if you had \[\sqrt[4]{-50}\], you cannot have that. so, that x=-50 would not be in the domain
what would the vertical asymptote be then?
vertical asymptote - the value of 'x' that the function approaches but never really gets there
now, you can have 0.0000000000001 and -0.00000000001 but not 0 so, you can approach x=0 but not get there
so, vertical asymptote -> x=0
okay so meaning that they horzontal/oblique asymptote is x?
no.. asymptote of f(x)... I just explained vertical.. similarly for hortizontal.. value of y but never gets there
so, we say that y "will" get there only at infinity... so, replace x by 1/u simplify, and put u=0. the value of f(x) you get will be your horizontal asymptote
oblique is that f(x) comes close to some line but never intercepts it.
well I graph the equation that isn't it a oblique asymptote?
no there are no oblique asymptotes here|dw:1363458832342:dw|
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