Question

Question #f43e1

Answer 1

It is `(h_(x))^-1 = 1/(3x^2 - 1)`

Explanation:

Inverse function of `h_(x) = 3x^2 - 1` is

`(h_(x))^-1 = 1/(3x^2 - 1)`

Answer 2

the inverse of the function is `sqrt[(x+1)/3]`

Explanation:

let `y=h(x)rArrx=h^-1(y)` but `h(x)=3x^2-1 rArry=3x^2-1rArrx^2=(y+1)/3`
`rArrx=sqrt[(y+1)/3]`
but `x=h^-1(y)rArrh^-1(y)=sqrt[(y+1)/3`
`rArrh^-1(x)=sqrt[(x+1)/3`

Answer 3

See below.

Explanation:

A function to have inverse, must be strictly increasing or strictly decreasing.

`h(x) = 3x^2-1`

is neither, so it does not have inverse.

NOTE:

Analyzing `h'(x) = 6x` we can observe that it is strictly decreasing for `x in (-oo,0)` because there `h'(x) lt 0` and strictly increasing for `x in (0,oo)` because there `h'(x) > 0`