Help please? :) [unit 2]

[lshak]

I feel so dumb lol but would anyone be able to help me with some of my holiday h/w?

My teacher sucks and I have no idea how to do these! haha they are probably really simple but range/domain has always confused me -_-









I wouldn't have a clue how to do these on my CAS either 

[pi]

4. Dom when the part inside the root is greater or equal to 0 (ie when the function exists). In this case,


B

5. Clue, the 'cusps' indicate that it is a magnitude graph. Looking a the y-int (which would be (0,-a) if the graph was not a magnitude), it must be option C.

6. Semi-circle. therefore E.

7. Inverse. You'd get D. Thanks moekamo, my bad

[b^3]

5. It is going to be a mod because of the all positive above the axis, so the orginal would be (x-root(a))(x+root(a))=x^2-a
so the equation is |x^2-a| so asnwer is c.

Edit: Beaten, Rohitpi, so fast

[moekamo]

5. that looks like x^2-a except that it is positive between -sqrt(a) and sqrt(a), so we have a |x^2-a| graph. so C

6. this is a the lower half of a semicircle of radius 2 and center at (0,0). hence the range is [-2,0]. so E

7. range of f inverse = domain of f. so ran = [1,3]. so D


EDIT: Beaten, except Rohitpi you got the range of f, not f inverse! im pretty sure its D

7. Same as above. You'd get C.

[pi]

EDIT: Beaten, except Rohitpi you got the range of f, not f inverse! im pretty sure its D

7. Same as above. You'd get C.

+1, I needed to expand my window Couldn't see the whole question

[pi]

^^^ For Q4, the graph has a 'closed circle', it exists at x=-3

woops

It happens.

[lshak]

Wow so fast haha thanks so much guys!

Uhmm what would this one be?

For which of the following would an inverse function exist regardless of the domain?
A. parabola
B. truncus
C. circle
D. hyperbola
E. absolute function graph

[moekamo]

hyperbola, if you draw a horizontal line anywhere it will only cross the function once, all the others have two crossing points.

[brightsky]

think about it logically, if you reflect a many-to-one function about the line y=x, then the resulting graph would have a one-to-many relation (i.e. if you put a vertical line through it, it will cross at more than one point), hence not being a "function". the question asks for an inverse FUNCTION, so only the functions with one-to-one relations work.

[lshak]

Ahh I get it, thanks so much!

Edit/  Okay what about this one?



For (a)...Is the domain just [-3,3] ?
For (c) is it  3 +or- squareroot k  (I just guessed )
And for (d) is it k>0 ??

....And I have no clue how to do (b) 

[hairypenis]

i think i got the same test as you! i haven't got my result back but i had no idea how to do that question ^^^either lol

[xZero]

a) [-3,3]
b) [-3+k,3+k]
c) +or- sqrt(9-k^2)
d) k>3

[taiga]

i think i got the same test as you! i haven't got my result back but i had no idea how to do that question ^^^either lol

That's because you're the same person? :S

I personally don't mind if you just post "bump" to get your question noticed.

PM me if I'm wrong, if not let me know which one of your (3) accounts you'd like to keep

You're not trying to do anything bad so obviously just take it as advice not to do it again

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This graph is a a circle with radius 3. centre at (0,k). hence the y values of the range are K+3 and k-3, the domain is from [-3,3]

edit: xZero has it my bad

[kamil9876]

Interesting, according to the rest of the world in question 7 f would not have an inverse (a technicality, need to change the codomain) but I guess in vce it does. Fun Fact.

[xZero]

Interesting, according to the rest of the world in question 7 f would not have an inverse (a technicality, need to change the codomain) but I guess in vce it does. Fun Fact.

Was trying to figure out why cant it have an inverse but failed miserably don't see anything wrong with R as the codomain, care to explain? thanks