What is the inverse of f(x)= 1/((x+4)^2) ?

The inverse of f(x)= 1/((x+4)^2) is sqrt(1/x)-4,-sqrt(1/x)-4

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inverse of f(x)= 1/((x+4)^2)

inverse

f (x) = \frac{1}{( x + 4 ) ^{2}}

\frac{1}{x} - 4, - \frac{1}{x} - 4

Solution steps

y = \frac{1}{( x + 4 ) ^{2}}

Replace

x

with

y

x = \frac{1}{( y + 4 ) ^{2}}

Solve for

y, x = \frac{1}{( y + 4 ) ^{2}}

\frac{1}{x} - 4, - \frac{1}{x} - 4

f (x) = \frac{5}{3} x^{3} - \frac{15}{2} x^{2}

x^{4} - 2 x^{3}

f (x) = x^{\frac{9}{2}} - 3 x^{2}

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

(- 3, 2), (- 3, - 2)

What is the inverse of f(x)= 1/((x+4)^2) ?
The inverse of f(x)= 1/((x+4)^2) is sqrt(1/x)-4,-sqrt(1/x)-4