Thread: Ahh help. Functions?

Select the true statement about the graph of f(x) = -2x^3 + 6x^2 + 18x - 18.

Then I'm given four possible answers

A. (-1, -31) is a relative minimum and (3, 32) is a relative maximum
B. (3, 32) is a relative minimum and (-1, -31) is a relative maximum
C. (-1, -28) is a relative minimum and (3, 36) is a relative maximum
D. (3,36) is a relative minimum and (-1, -28) is a relative maximum

I believe it's C. I'm also thinking it could be D, if I got the numbers mixed up.


and also, this one has me stuck as well. I haven't the slightest clue what it's even asking.

Which of the following functions is neither odd nor even?
A. x^9 + 3x^7 + 6x
B. x^5 + 3x
C. 6x^2 - (x^2) + 3
D. x^2 + 3x + 3 + (x - 3)

Originally Posted by amandaa

Select the true statement about the graph of f(x) = -2x^3 + 6x^2 + 18x - 18.

Then I'm given four possible answers

A. (-1, -31) is a relative minimum and (3, 32) is a relative maximum
B. (3, 32) is a relative minimum and (-1, -31) is a relative maximum
C. (-1, -28) is a relative minimum and (3, 36) is a relative maximum
D. (3,36) is a relative minimum and (-1, -28) is a relative maximum

I believe it's C. I'm also thinking it could be D, if I got the numbers mixed up.

as for this why do u think it's C or D ? when function is max or min ? and you should not have any problem with that if you don't know say but you are on the right track it's correct answer (one of those)

Originally Posted by amandaa

Which of the following functions is neither odd nor even?
A. x^9 + 3x^7 + 6x
B. x^5 + 3x
C. 6x^2 - (x^2) + 3
D. x^2 + 3x + 3 + (x - 3)

as for this
odd function is function that satisfy and even function is meaning that you need to put (-x) in every place of x in function and see which conditions does it satisfy


P.S. you need function that does not satisfy any of that two conditions

Originally Posted by yeKciM

as for this why do u think it's C or D ? when function is max or min ? and you should not have any problem with that if you don't know say but you are on the right track it's correct answer (one of those)




as for this
odd function is function that satisfy and even function is meaning when you put in you function instead of x you put (-x) in places of every x that is there in function, and see which condition does it satisfy


P.S. you need function that does not satisfy any of that two conditions

Well when I plug in -1,-28 it fits the equation, and so does 3,36. I do think it's C since -1,-28 makes more sense to be the minimum.

As for the 2nd question, I already knew what you said, but I don't know how to set it up, or even solve it D:

yes it's C that first question .... to test is it min or max use second derivate of function

as for second ... i'll show you one

A:


so it's odd function okay ?


hint: it's easy to conclude that if (-x) is to power of 2,4,6,8,... it will stay x (when you put (-x) to see ) and if it's to power of 1,3,5,7,... when put (-x) it will stay (-x) ... so if your function have all x at power of 2,4,6,8... it will be even function, but if they are all 1,3,5,7,... it is odd function, so if they are combined than function is neither odd neither even function

Thanks!
Okay so, I still don't understand it lol. I'm so horrible at this. I tried the second one, but I don't think I did it right.

B: f(x) = x+5+3x
f(-x) = -x+5+3(-x)
= -x+5-3x
=5-3x^2

I don't think that's right. And if it is, I can't tell if it's odd or not

Originally Posted by amandaa

Select the true statement about the graph of f(x) = -2x^3 + 6x^2 + 18x - 18.

Then I'm given four possible answers

A. (-1, -31) is a relative minimum and (3, 32) is a relative maximum
B. (3, 32) is a relative minimum and (-1, -31) is a relative maximum
C. (-1, -28) is a relative minimum and (3, 36) is a relative maximum
D. (3,36) is a relative minimum and (-1, -28) is a relative maximum

I believe it's C. I'm also thinking it could be D, if I got the numbers mixed up.

how have you been taught to find extrema of a function?

Originally Posted by amandaa

Thanks!
Okay so, I still don't understand it lol. I'm so horrible at this. I tried the second one, but I don't think I did it right.

B: f(x) = x+5+3x
f(-x) = -x+5+3(-x)
= -x+5-3x
=5-3x^2

I don't think that's right. And if it is, I can't tell if it's odd or not

let's try this




which is odd and which is even ?

Originally Posted by skeeter

how have you been taught to find extrema of a function?

You know, I honestly don't remember. I haven't taken a math class in awhile, so I'm really struggling with all of this.

Originally Posted by yeKciM

let's try this




which is odd and which is even ?

sorry for double posting! I don't get what you're asking lol, and that made my confusion worse

Originally Posted by amandaa

You know, I honestly don't remember. I haven't taken a math class in awhile, so I'm really struggling with all of this.

do you know how to find derivation of the function ?


okay
let's look at that function to see is it odd or even function we have to put (-x) in place of x, of that function... so:


and because it's at the power of 2 is's like (-2)*(-2) = +4 so we have that

so that means that function f(-x) is equal function with f(x)



because those two are equal means that function f(x) is even function

Originally Posted by yeKciM

do you know how to find derivation of the function ?

Nope! I'm not getting any of this, so I'm finished. Thanks for the help though.

Originally Posted by amandaa

Nope! I'm not getting any of this, so I'm finished. Thanks for the help though.

sorry... i tired maybe it's just me... someone else could probably explain it to you much better
when doing math it's very important to be persistent, and don't give up easy

Originally Posted by yeKciM

sorry... i tired maybe it's just me... someone else could probably explain it to you much better

This is in the precalculus section, so it's unlikely (though not impossible) that the poster of the thread knows about derivatives. There are other ways to find extrema. skeeter seemed to be getting at this. However, if amandaa isn't sure what method they were taught, it's very difficult to give appropriate help.