What is the inverse function of #y=3x^2-5#?

Question

Madelyn Anderson · Accepted Answer

#x=sqrt((y+5)/3)#

To find the inverse of a function expressed as #color(white)("XXX")y= # an expression in #x# rearrange the expression into the form: #color(white)("XXX")x= # an expression in #y#

Given #color(white)("XXX")y=3x^2-5#

Switch sides and add #5# to both sides #color(white)("XXX")3x^2=y+5#

Divide both sides by #3# #color(white)("XXX")x^2=(y+5)/3#

Take the square root of both sides #color(white)("XXX")x= +-sqrt((y+5)/3)#

Since we were asked for an inverse function, I've eliminated the non-primary root (a function can not have two values for a single argument value).

Therefore #color(white)("XXX")x=sqrt((y+5)/3)#

Lily Anderson · Answer

With the default domain, this function has no inverse since it is not one-to-one, but read on...

Let #f(x) = 3x^2-5#. The implicit domain of #f# is the set of values for which it is defined. By convention, since the variable name is #x# rather than #z#, we're talking about #RR# rather than #CC# and #f# is well defined for the whole of #RR#. However, #f# is not one-to-one in that #f(-x) = f(x)# for all #x in RR#, so #f# has no well-defined inverse on its default domain. We can try to find an inverse as follows: Let #y = f(x) = 3x^2-5#. Adding #5# to both ends we get: #y+5 = 3x^2# Dividing both sides by #3# we get: #x^2 = (y+5)/3# Hence #x = +-sqrt((y+5)/3)# This is not uniquely defined, so does not define a function, unless... If we restrict the domain of #f# to #[0, oo)# then #f# is one-to-one and has inverse: #f^(-1)(y) = sqrt((y+5)/3)# Alternatively, if we restrict the domain of #f# to #(-oo, 0]#, then #f# is one-to-one with inverse: #f^(-1)(y) = -sqrt((y+5)/3)# Interestingly, if we define #g(y) = sqrt((y+5)/3)#, then #g(y)# has an inverse function #g^(-1)(x) = f(x)#. If we define #h(y) = -sqrt((y+5)/3)#, then #h(y)# has an inverse function #h^(-1)(x) = f(x)#. If we define: #k(y) = { (sqrt((y+5)/3), "if " y " is rational"), (-sqrt((y+5)/3), "if " y " is irrational") :}# then #k(y)# has an inverse function #k^(-1)(x) = f(x)# The implicit domain for #g(y), h(y) and k(y)# is #[-5, oo)# since we require #y >= -5# in order that the square root has a Real value.

Madelyn Anderson · Answer

undefined To find the inverse function of $ y = 3x^2 - 5 $, first, interchange $ x $ and $ y $ to get $ x = 3y^2 - 5 $. Then, solve for $ y $ in terms of $ x $.

$$ x = 3y^2 - 5 $$
$$ x + 5 = 3y^2 $$
$$ \frac{{x + 5}}{3} = y^2 $$
$$ y = \pm \sqrt{\frac{{x + 5}}{3}} $$

Therefore, the inverse function is:
$$ f^{-1}(x) = \pm \sqrt{\frac{{x + 5}}{3}} $$