What is the inverse of h(x)=-x^2+6,x>= 0 ?

The inverse of h(x)=-x^2+6,x>= 0 is sqrt(-x+6),-sqrt(-x+6)

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inverse of h(x)=-x^2+6,x>= 0

inverse

h (x) = - x^{2} + 6, x \geq 0

- x + 6, - - x + 6

Solution steps

y = - x^{2} + 6

Replace

x

with

y

x = - y^{2} + 6

Solve for

y, x = - y^{2} + 6

- x + 6, - - x + 6

f (x) = \frac{9 x}{8 x - 3}

f (x) = 16 x^{2} - 24 x + 6

f (x) = x^{2} + 2 x - 3, x \leq 2

\frac{x - 4}{x + 7}

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

What is the inverse of h(x)=-x^2+6,x>= 0 ?
The inverse of h(x)=-x^2+6,x>= 0 is sqrt(-x+6),-sqrt(-x+6)