Question

Include work too please (ALGEBRA 2)

Answer 1

Explanation:

When the inverse was being calculated , they took the square root of one side and didn't +- the root on the other side of the equation

so NEITHER is correct

f^-1 SHOULD be f^-1(x) = +- 4 sqrt(x+1) +3

then domain is x>= -1 ( this is correct) BUT range is +- infinity

Find the original function :

change x's and y's ....then solve for y

x =+- 4 sqrt(y+1) + 3

+- (x-3)/4 = sqrt (y+1) square both sides

1/16 (x-3)^2 = y+1

1/16 ( x-3)^2 -1 = f(x)

Here is graph of f(x) and f^-1(x) :

Answer 2

solution:-

• Let f^-1(x) = y

• so, y = 4 √(x+1) + 3

• y - 3 = 4√(x+1)

• y-3/4 = √x+1

• squaring both side ,

• (y-3/4)² = x+1

• ( y-3 )² ≥ 0

• hence , y ≥ 3

• and x + 1 ≥ 0

• since, x ≥ -1

• Since student a is right .

• similarly in 2nd

• (y-3/ -4)² = x + 1

• (3-y /4)² = x + 1

• 3- y ≥ 0

• 3 ≥ y

• x≥-1

here both students are right .

1. now for (b) equation of f(x)

We have got, (y-3)²/16 = x +1

change y into x and x into f(x)

we got , (x -3)² /16 -1 = f(x)

f(x) = (x-3)²/16 - 1 is required equation.