What is the inverse of f(x)=(x-7)^2+1,x<= 7 ?

The inverse of f(x)=(x-7)^2+1,x<= 7 is sqrt(x-1)+7,-sqrt(x-1)+7

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inverse of f(x)=(x-7)^2+1,x<= 7

inverse

f (x) = (x - 7)^{2} + 1, x \leq 7

x - 1 + 7, - x - 1 + 7

Solution steps

y = (x - 7)^{2} + 1

Replace

x

with

y

x = (y - 7)^{2} + 1

Solve for

y, x = (y - 7)^{2} + 1

x - 1 + 7, - x - 1 + 7

\frac{5 x}{x + 7}

f (x) = \frac{( 3 x + 5 )}{2}

\frac{x ^{2} + 1}{x ^{4} + 3 x ^{2} + 1}

6 x^{3}

4 + x^{2}

What is the inverse of f(x)=(x-7)^2+1,x<= 7 ?
The inverse of f(x)=(x-7)^2+1,x<= 7 is sqrt(x-1)+7,-sqrt(x-1)+7