What is the inverse of f(x)=(x+4)^2,x>= 0 ?

The inverse of f(x)=(x+4)^2,x>= 0 is sqrt(x)-4,-sqrt(x)-4

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inverse of f(x)=(x+4)^2,x>= 0

inverse

f (x) = (x + 4)^{2}, x \geq 0

x - 4, - x - 4

Solution steps

y = (x + 4)^{2}

Replace

x

with

y

x = (y + 4)^{2}

Solve for

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

x - 4, - x - 4

x^{2} + 6 x + 9

\frac{x ^{2} + 3 x - 10}{x ^{2} - 4}

f (x) = \frac{1}{1 + x ^{4}}

f (x) = 3.5 (2.2)^{x}

What is the inverse of f(x)=(x+4)^2,x>= 0 ?
The inverse of f(x)=(x+4)^2,x>= 0 is sqrt(x)-4,-sqrt(x)-4