What is the inverse of h(x)=(-10x+17)^2,x<= 17/10 ?

The inverse of h(x)=(-10x+17)^2,x<= 17/10 is -(sqrt(x)-17)/(10),-(-sqrt(x)-17)/(10)

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inverse of h(x)=(-10x+17)^2,x<= 17/10

inverse

h (x) = (- 10 x + 17)^{2}, x \leq \frac{17}{10}

- \frac{x - 17}{10}, - \frac{- x - 17}{10}

Solution steps

y = (- 10 x + 17)^{2}

Replace

x

with

y

x = (- 10 y + 17)^{2}

Solve for

y, x = (- 10 y + 17)^{2}

- \frac{x - 17}{10}, - \frac{- x - 17}{10}

What is the inverse of h(x)=(-10x+17)^2,x<= 17/10 ?
The inverse of h(x)=(-10x+17)^2,x<= 17/10 is -(sqrt(x)-17)/(10),-(-sqrt(x)-17)/(10)