What is the inverse of f(x)=3(x+4)^2-2,x<= 4 ?

The inverse of f(x)=3(x+4)^2-2,x<= 4 is sqrt((x+2)/3)-4,-sqrt((x+2)/3)-4

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inverse of f(x)=3(x+4)^2-2,x<= 4

inverse

f (x) = 3 (x + 4)^{2} - 2, x \leq 4

\frac{x + 2}{3} - 4, - \frac{x + 2}{3} - 4

Solution steps

y = 3 (x + 4)^{2} - 2

Replace

x

with

y

x = 3 (y + 4)^{2} - 2

Solve for

y, x = 3 (y + 4)^{2} - 2

\frac{x + 2}{3} - 4, - \frac{x + 2}{3} - 4

tan (- 2)

f (x) = 20000 - (2 \cdot x)

f (x) = \frac{( 2 x - 1 )}{( x + 2 )}

\frac{x + 5}{x - 6}

f (x) = 3 x + 11

What is the inverse of f(x)=3(x+4)^2-2,x<= 4 ?
The inverse of f(x)=3(x+4)^2-2,x<= 4 is sqrt((x+2)/3)-4,-sqrt((x+2)/3)-4