What is the inverse of f(x)=-5x^2-9,x>= 0 ?

The inverse of f(x)=-5x^2-9,x>= 0 is sqrt(-(x+9)/5),-sqrt(-(x+9)/5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)=-5x^2-9,x>= 0

inverse

f (x) = - 5 x^{2} - 9, x \geq 0

- \frac{x + 9}{5}, - - \frac{x + 9}{5}

Solution steps

y = - 5 x^{2} - 9

Replace

x

with

y

x = - 5 y^{2} - 9

Solve for

y, x = - 5 y^{2} - 9

- \frac{x + 9}{5}, - - \frac{x + 9}{5}

f (x) = x^{2} - 16 x + 4

f (x) = \frac{5}{7 x - 8}

f (x) = 0.2 sin (x)

f (x) = \frac{3 x - 4}{5}

f (x) = \frac{x ^{4}}{16}

What is the inverse of f(x)=-5x^2-9,x>= 0 ?
The inverse of f(x)=-5x^2-9,x>= 0 is sqrt(-(x+9)/5),-sqrt(-(x+9)/5)