Question

Find the range of the function: \(\Large y = \dfrac{x^2}{16x - 1} \)

Answer 1

@dpaInc

Answer 2

since the degree of the numerator is 1 more than the degree of the denominator, you'll have a slant asymptote.
this means the range will be a subset of all reals.

Answer 3

are you familiar with finding local/relative extrema?

Answer 4

Kind of, I guess.

Answer 5

Ok, maybe not.

Answer 6

ok find the derivative, y' = ???

Answer 7

do you know about derivatives?

Answer 8

Applying quadrant rule: I got \(\Large y' = \dfrac{2x(16x - 1) - 16x^2 }{(16x - 1)^2} \).

Answer 9

close...
this is what i got: \(\large y'=\frac{2x(8x-1)}{(16x-1)^2} \)

Answer 10

Which can be reduced to \( \Large y' = \dfrac{16x^2 - 2x}{(16x^2-1)^2} \)

It's the same. Now what?

Answer 11

find the critical numbers....

Answer 12

no it's not the same... look at what you gave as y' compared to what you have in the latter post... they're not the same.

Answer 13

no matter, your last post for the derivative is correct now...

Answer 14

now find critical numbers and determine whether they are local max or local mins...

Answer 15

if you use the factored form of the derivative, you'll get x=0 and x=1/8.

so the critical numbers will be x=0, x=1/8
by doing either first derivative test or second derivative test you'll find that x=0 is a local max; and x=1/8 is a local min.

Answer 16

find the y values at these critical numbers and you'll get the boundaries for the range.