+0

#9 ​Help.

+1
377
2

#9 Help.

Feb 12, 2018

#2
+1

#10, why not?

According to the original relation, $$f(x)=\sqrt{x-4}$$. There are a few steps that need to be done in order to find the inverse.

#1) Convert to y=-notation. I find that it is easier when working with this notation. This is quite a simple step, wouldn't you agree?

$$f(x)=\sqrt{x-4}\Rightarrow y=\sqrt{x-4}$$

#2) Now, replace all instances of a "y" with an "x," and replace all instances of an "x" with a "y." This is not a difficult step, either, as you might imagine.

$$y=\sqrt{x-4}\Rightarrow x=\sqrt{y-4}$$

#3) Solve for y. This step can range in difficulty. In this case, it is quite a simple step.

 $$x=\sqrt{y-4}$$ Cube both sides. $$x^3=y-4$$ Add 4 to both sides to isolate y. $$y=x^3+4$$

#4) Convert back to function notation since the original problem was given in function notation. This is quite simple, too.

$$y=x^3+4\Rightarrow f^{-1}(x)=x^3+4$$

Feb 14, 2018

#1
0

Anyone for number 10?

Feb 12, 2018
#2
+1

#10, why not?

According to the original relation, $$f(x)=\sqrt{x-4}$$. There are a few steps that need to be done in order to find the inverse.

#1) Convert to y=-notation. I find that it is easier when working with this notation. This is quite a simple step, wouldn't you agree?

$$f(x)=\sqrt{x-4}\Rightarrow y=\sqrt{x-4}$$

#2) Now, replace all instances of a "y" with an "x," and replace all instances of an "x" with a "y." This is not a difficult step, either, as you might imagine.

$$y=\sqrt{x-4}\Rightarrow x=\sqrt{y-4}$$

#3) Solve for y. This step can range in difficulty. In this case, it is quite a simple step.

 $$x=\sqrt{y-4}$$ Cube both sides. $$x^3=y-4$$ Add 4 to both sides to isolate y. $$y=x^3+4$$

#4) Convert back to function notation since the original problem was given in function notation. This is quite simple, too.

$$y=x^3+4\Rightarrow f^{-1}(x)=x^3+4$$