What is the inverse of f(x)=x^2-6x,3<= x<= 5 ?

The inverse of f(x)=x^2-6x,3<= x<= 5 is 3+sqrt(x+9),3-sqrt(x+9)

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inverse of f(x)=x^2-6x,3<= x<= 5

inverse

f (x) = x^{2} - 6 x, 3 \leq x \leq 5

3 + x + 9, 3 - x + 9

Solution steps

y = x^{2} - 6 x

Replace

x

with

y

x = y^{2} - 6 y

Solve for

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

3 + x + 9, 3 - x + 9

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

f (x) = x^{4} - 8 x^{2} + 7

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

\frac{s + 1}{s - 2}

f (x) = 3 cos (2 x)

What is the inverse of f(x)=x^2-6x,3<= x<= 5 ?
The inverse of f(x)=x^2-6x,3<= x<= 5 is 3+sqrt(x+9),3-sqrt(x+9)