Given: f(x)= sqrt(4-x)
and h(x)= x^2

Find g, the inverse of f. GIve the domain and range for both f and g, sketch the graphs of f and g on the same set of axes.

Question

Given: f(x)= sqrt(4-x)
 and h(x)= x^2 

Find g, the inverse of f. GIve the domain and range for both f and g, sketch the graphs of f and g on the same set of axes.

calculusfunctions · Answer

Do you how to find the inverse of a function, algebraically?

anonymous · Answer

isnt it the reciprocal of it ? and opposite sign

calculusfunctions · Answer

No! I was just writing the lesson for you so give a few minutes to type it up. I will explain everything. Alright?

anonymous · Answer

ok

calculusfunctions · Answer

The inverse relation is the reflection in the line  y = x.

To find the inverse of a function:

i). rewrite y = f(x)
ii). Interchange x and y
iii). solve for y
iv). write the inverse as$$f ^{-1}(x)=...$$where$$f ^{-1}\neq \frac{ 1 }{ f }$$That is a symbol used to signify an inverse of a function and is not in exponent, even though it's used in the place of an exponent. Understood?

Example:
Find the inverse of$$f(x)=4(x -1)^{2}-2$$Following the steps above we write$$y =4(x -1)^{2}-2$$
$$x =4(y -1)^{2}-2$$
$$x +2=4(y -1)^{2}$$
$$\frac{ x +2 }{ 4 }=(y -1)^{2}$$
$$\pm \sqrt{\frac{ x +2 }{ 4 }}=y -1$$
$$y =1\pm \frac{ \sqrt{x +2} }{ 2 }$$
$$y =\frac{ 2 }{ 2 }\pm \frac{ \sqrt{x +2} }{ 2 }$$Therefore$$f ^{-1}(x)=\frac{ 2\pm \sqrt{(x +2)} }{ 2 }$$

Do you understand? If yes then do your question and I'll check your work after you're done.

anonymous · Answer

In the last steps I don't understand why you changed 1 to 2/2

calculusfunctions · Answer

OH! sorry I was just writing more notes for you when I noticed your question.

I changed the 1 into 2/2 because I wanted to simplify so that I can put everything over a single denominator. That's all.

anonymous · Answer

ooh okay thanks! uhm let me try with mine

anonymous · Answer

for my question i got -x+4=y

calculusfunctions · Answer

Because an inverse and the given relation are inverse relations, the Domain of the one is the range of the other, and the range of one is the domain of the other. 

Also note that if the given relation is a function, doesn't guarantee that the inverse is a function. The inverse of a function is also a function iff the given function is a one-to-one function. A graph of a relation is function if it passes the vertical line test (any vertical line drawn through the graph will intercept the graph at no more than one point). A graph of a relation is a one-to-one function iff it passes the horizontal line test (any horizontal line drawn through the graph intercepts the graph at no more than one point). If it passes both the vertical and a horizontal line test then the relation is both a function and a one-to-one function. However, if the inverse of the first function is to be a function than the first function only has to be a one-to-one function (pass the horizontal line test).

calculusfunctions · Answer

Can you also make the effort to show me all your steps please just like I made the effort to type an entire lesson for you? Thanks!

calculusfunctions · Answer

You answer is incorrect anyway. So could you show me your work so that I can point out where you went wrong and teach you?

anonymous · Answer

$$f(x)=\sqrt{4-x} $$
$$y=\sqrt{4-x}$$
$$x=\sqrt{4-y}$$
$$(x)^2=(\sqrt{4-x})^2$$
$$x^2-4=-y$$
$$-x^2+4=y$$

anonymous · Answer

$$f ^{-1}(x)= -x^2+4$$

anonymous · Answer

@calculusfunctions

calculusfunctions · Answer

Yes, actually that's it!! Great but again no work.

anonymous · Answer

I showed my work

anonymous · Answer

Its above my final answer.

calculusfunctions · Answer

OOPS! Sorry, I didn't see that. That's Perfect!! except in the third step after the question is written you have x on both sides and have inadvertently omitted y. I know that wasn't intentional but just be careful on tests and assignments. Otherwise, excellent work!!

anonymous · Answer

Thanks! I have one other question you said "Because an inverse and the given relation are inverse relations, the Domain of the one is the range of the other, and the range of one is the domain of the other. " I dont understand that statement.. To find the domain do i set $$f ^{-1}(x)= -x^2+4$$ ----> $$-x^2+4\ge 0$$

anonymous · Answer

@calculusfunctions

calculusfunctions · Answer

I meant that if you know the domain of the given function then that is the range of the inverse function and the range of the given function is the domain of the inverse function.

Example:$$f(x)=4(x -1)^{2}-2$$
$$Domain:\left\{ x \inℝ \right\}$$and $$Range:\left\{ y \inℝ|y \ge -2 \right\}$$Then the domain and range of the inverse are as follows:$$Domain:\left\{ x \inℝ|x \ge -2 \right\}$$and$$Range:\left\{ y \in ℝ\right\}$$

Understand? I have to go now but will probably be back in a few hours. If yo need me for anything then message me if you see me on line.