Mathematics: Find the inverse of “f” algebraically, please!

Find the inverse of “f” algebraically, please!

Ответ:

briarkaltvedt

13.02.2022

Mathematics

$f^{-1}(x) = -\sqrt{36-x^2}$

====================================================

Here are the steps to find the inverse

$f(x) = \sqrt{36-x^2}\\\\y = \sqrt{36-x^2}\\\\x = \sqrt{36-y^2} \ \text{ swap x and y; solve for y}\\\\x^2 = 36-y^2\\\\x^2+y^2 = 36\\\\y^2 = 36-x^2\\\\y = \sqrt{36-x^2}\\\\f^{-1}(x) = \sqrt{36-x^2}\\\\$

Now this would be the answer if our original domain was $0 \le x \le 6$

However, that's not the case and instead our domain is $-6 \le x \le 0$

Reflecting any (x,y) point in this domain restriction, over the the line y = x, will have us land on a point with a negative y value. This is because the inverse will swap the x and y values.

Therefore, the inverse function must ultimately be negative.

Currently $f^{-1}(x) = \sqrt{36-x^2}$ is positive since the result of any square root is never negative.

To fix this, we stick a negative out front to get the final answer being $f^{-1}(x) = -\sqrt{36-x^2}$

0,0(0 оценок)

Find the inverse of “f” algebraically, please!

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question