What is the inverse of f(x)=5x^2-4,x>= 0 ?

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

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inverse of f(x)=5x^2-4,x>= 0

inverse

f (x) = 5 x^{2} - 4, x \geq 0

\frac{x + 4}{5}, - \frac{x + 4}{5}

Solution steps

y = 5 x^{2} - 4

Replace

x

with

y

x = 5 y^{2} - 4

Solve for

y, x = 5 y^{2} - 4

\frac{x + 4}{5}, - \frac{x + 4}{5}

f (x) = 1 + \frac{4}{5} x

f (x) = \frac{5 x + 9}{- 2 x + 4}

f (x) = \frac{2 x - 1}{x - 2}

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

f (x) = 3^{x + 2} - 6

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