What is the inverse of f(x)=x^2+8x,x>=-4 ?

The inverse of f(x)=x^2+8x,x>=-4 is -4+sqrt(x+16),-sqrt(x+16)-4

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inverse of f(x)=x^2+8x,x>=-4

inverse

f (x) = x^{2} + 8 x, x \geq - 4

- 4 + x + 16, - x + 16 - 4

Solution steps

y = x^{2} + 8 x

Replace

x

with

y

x = y^{2} + 8 y

Solve for

y, x = y^{2} + 8 y

- 4 + x + 16, - x + 16 - 4

\frac{7}{5 x + 3}

\frac{x ^{2}}{- 2 + x}

f (x) = \frac{x}{x + 8}

y = 2 x + 4 (4.4)

f (x) = 4 sec (6 x - 2 π) - 12

What is the inverse of f(x)=x^2+8x,x>=-4 ?
The inverse of f(x)=x^2+8x,x>=-4 is -4+sqrt(x+16),-sqrt(x+16)-4