Home > Precalculus > Introduction to Twelve Basic Functions

How do you find #f^-1(x)# given #f(x)=3/(x^2+2x)#?

Answer 1

Serenity Jones

The answer is #=-1+-sqrt(x(x+3))/x#

Let #y=3/(x^2+2x)#

Then,

#y(x^2+2x)=3#

#yx^2+2yx-3=0#

Comparing this equation to

#ax^2+bx+c=0#

Calculating the discriminant

#Delta=b^2-4ac=(2y)^2-4*y+(-3)#

#=4y^2+12y=4y(y+3)#

So,

#x=(-b+-sqrtDelta)/(2a)#

#x=((-2y)+-sqrt(4y(y+3)))/(2y)#

#x=-1+-sqrt(y(y+3))/y#

Interchanging #x# and #y#

#y=-1+-sqrt(x(x+3))/x#

Therefore,

#f^-1(x)=-1+-sqrt(x(x+3))/x#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Liam Brennan

To find the inverse function ( f^{-1}(x) ) given ( f(x) = \frac{3}{x^2 + 2x} ), follow these steps:

Replace ( f(x) ) with ( y ): ( y = \frac{3}{x^2 + 2x} ).
Swap the variables ( x ) and ( y ): ( x = \frac{3}{y^2 + 2y} ).
Solve the equation for ( y ).
Replace ( y ) with ( f^{-1}(x) ) to express the inverse function in terms of ( x ).

Following these steps:

( y = \frac{3}{x^2 + 2x} ).
( x = \frac{3}{y^2 + 2y} ).
Solve for ( y ):
- ( x(y^2 + 2y) = 3 )
- ( xy^2 + 2xy - 3 = 0 )
- Solve this quadratic equation for ( y ).
Once you find the expression for ( y ), replace ( y ) with ( f^{-1}(x) ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.