Question #f43e1

3 Answers
Nov 3, 2017

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)#

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#

Nov 3, 2017

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#