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}

2 x + 8^{2}

f (x) = \frac{x ^{7}}{8}

\frac{1}{2} \cdot e^{3 - x} + 2

f (x) = (x + 2)^{2} - 7

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)