If f(x)=3x^2+1 and g(x)=2x+2 find a) g^-1(x), the inverse of g(x) b) (fog^-1)(2)

Question

Help me please I need to pass math and this is due tomorrow

Answer 1

Hi Hannah,

 

If g(x) = 2x + 2, this means it takes an input value, doubles it, and then adds 2.  The inverse will do the opposite, in the opposite order.  

 

So the inverse will first do the opposite of adding 2, so it subtracts 2, or x - 2.  Then it does the opposite of multiplying, by dividing by 2, so (x-2) / 2.

 

So g^-1(x)= (x-2) / 2.   

 

To find (fog^-1)(2), you start by putting 2 into g inverse for x .  So  g^-1(2) = (2-2) /2 , so zero.  Then, we put 0 into f, so f(0) = 3(0)² + 1, so (fog^-1)(2) = 1.  Hope that helped.

Answer 2

Given f(x) = 3x^2 + 1 and g(x) = 2x + 2

a) g^-1(x), the inverse of g(x)
Let’s follow a few steps for inverse functions:
Put y for the g(x) expression,
Switch x for y
Solve for y as the inverse function

Thus y = 2x + 2
Switching x for y,
X = 2y + 2
2y = x – 2
Y = (x-2)/2
So g^-1(x) = (x-2)/2


b) (fog^-1)(2) means substituting g inverse (g^-1) into f(x) when X = 2
Thus, fog^-1(x); x = 2
fog^-1(x) = 3((x-2)/2)^2) + 1
when x = 2
fog^-1(2) = 3((2-2)/2)^2) + 1
= 0 +1
= 1