Question

Choose the inverse of y = x2 – 10x.

Answer 1

we have

$y=x^{2}-10x$

Completing the square

$y+5^{2}=x^{2}-10x+5^{2}$

$y+25=x^{2}-10x+25$

$y+25=(x-5)^{2}$

Swap  x and y

$x+25=(y-5)^{2}$

Solve for y

$(y-5)^{2}=x+25$

Square root both sides

$(y-5)=(+/-)\sqrt{x+25}$

$y=(+/-)\sqrt{x+25}+5$

Let

$f(x)^{-1} =y$

$f(x)^{-1}=(+/-)\sqrt{x+25}+5$

therefore

the answer is

The inverse is equal to

$f(x)^{-1}=(+/-)\sqrt{x+25}+5$

Answer 2

The inverse of a function, is its opposite.

The inverse function is: $\mathbf{f^{-1}(x) = 5 \pm \sqrt{x+25} }$

The function is given as:

$\mathbf{y = x^2 - 10x}$

Swap y and x

$\mathbf{x = y^2 - 10y}$

Complete the square on the above equation

$\mathbf{x+ (-\frac{10}{2})^2 = y^2 - 10y + (-\frac{10}{2})^2}$

$\mathbf{x+25 = y^2 - 10y + 25}$

Express the right-hand side as a perfect square

$\mathbf{x+25 = (y - 5)^2}$

Take square roots of both sides

$\mathbf{\pm \sqrt{x+25} = y - 5}$

Add 5 to both sides

$\mathbf{5 \pm \sqrt{x+25} = y }$

Rewrite as:

$\mathbf{y = 5 \pm \sqrt{x+25} }$

So, we have:

$\mathbf{f^{-1}(x) = 5 \pm \sqrt{x+25} }$

Hence, the inverse function is: $\mathbf{f^{-1}(x) = 5 \pm \sqrt{x+25} }$

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