What is the inverse of f(x)=7x^2+8,x>= 0 ?

The inverse of f(x)=7x^2+8,x>= 0 is sqrt((x-8)/7),-sqrt((x-8)/7)

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inverse of f(x)=7x^2+8,x>= 0

inverse

f (x) = 7 x^{2} + 8, x \geq 0

\frac{x - 8}{7}, - \frac{x - 8}{7}

Solution steps

y = 7 x^{2} + 8

Replace

x

with

y

x = 7 y^{2} + 8

Solve for

y, x = 7 y^{2} + 8

\frac{x - 8}{7}, - \frac{x - 8}{7}

\frac{x ^{3}}{4}

f (x) = \frac{2 x + 5}{4}

(7 - 14 x)^{\frac{1}{2}}

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

x^{2} + 3 x - 2

What is the inverse of f(x)=7x^2+8,x>= 0 ?
The inverse of f(x)=7x^2+8,x>= 0 is sqrt((x-8)/7),-sqrt((x-8)/7)