Question

Let
f (x)= 10x/(x2 + 81)

Find the largest interval containing 0 on which f has an inverse. Choose the correct interval from the list and enter the endpoint values in the appropriate blanks.Enter DNE in any unused answer blanks.

The largest interval is
(-infinity, infinity)
(-infinity, a)
(-infinity, a]
(a, infinity)
[a, infinity)
(a, b)
(a, b]
[a, b)
[a, b]


a =
b =

Answer 1

f (x)= 10x/(x^2 + 81)=y
10x=y*x^2 + 81y

y*x^2 -10x+ 81y=0
x=10/2y+(-)sqrt (100-4*81y^2)/2y

so 100-4*81*y^2>0
y^2<100/4*81
-10/18<y<10/18
y belongs to [-5/9,0) to (0,5/9]

Answer 2

10x/(x^2 + 81)=y
10x=y(x^2) + 81y

y(x^2) -10x+ 81y=0
x=10/2y+(-)sqrt (100-4(81)y^2)/2y

x is real , therefore discriminant is greater than or equal to 0 .

100-4(81)(y^2)>=0
y^2<=100/4(81)
-10/18<=y<=10/18
y is in [-5/9,5/9]

i.e , a= -5/9 and b= 5/9

Answer 3

(-infinity, infinity)

since f(x) exist on this interval and is one-one function