What is the inverse of h(x)=(8-x^2)/5 ?

The inverse of h(x)=(8-x^2)/5 is sqrt(-5x+8),-sqrt(-5x+8)

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inverse of h(x)=(8-x^2)/5

inverse

h (x) = \frac{8 - x ^{2}}{5}

- 5 x + 8, - - 5 x + 8

Solution steps

y = \frac{8 - x ^{2}}{5}

Replace

x

with

y

x = \frac{8 - y ^{2}}{5}

Solve for

y, x = \frac{8 - y ^{2}}{5}

- 5 x + 8, - - 5 x + 8

f (x) = \frac{1}{x ^{3} + 2}

f (x) = \frac{x + 6}{4}

f (x) = - ln (2 x + 3)

y = x + \frac{1}{x}

arccos (\frac{3}{2})

What is the inverse of h(x)=(8-x^2)/5 ?
The inverse of h(x)=(8-x^2)/5 is sqrt(-5x+8),-sqrt(-5x+8)