What is the inverse of f(x)=sqrt(x^2+5x),x>0 ?

The inverse of f(x)=sqrt(x^2+5x),x>0 is (-5+sqrt(25+4x^2))/2 ,(-5-sqrt(25+4x^2))/2

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inverse of f(x)=sqrt(x^2+5x),x>0

inverse

f (x) = x^{2} + 5 x, x > 0

\frac{- 5 + 25 + 4 x ^{2}}{2}, \frac{- 5 - 25 + 4 x ^{2}}{2}

Solution steps

y = x^{2} + 5 x

Replace

x

with

y

x = y^{2} + 5 y

Solve for

y, x = y^{2} + 5 y

\frac{- 5 + 25 + 4 x ^{2}}{2}, \frac{- 5 - 25 + 4 x ^{2}}{2}

2 x - y = 6

m = 1, (- 4, 7)

(0.4) (4)

f (x) = \frac{5}{7 x}

f (x) = 5 + 2 e^{x}

What is the inverse of f(x)=sqrt(x^2+5x),x>0 ?
The inverse of f(x)=sqrt(x^2+5x),x>0 is (-5+sqrt(25+4x^2))/2 ,(-5-sqrt(25+4x^2))/2