Giveaway One lucky person will win 10 million

[xxDianaxx]

Solve this math problem cause I love maths find f(x) x3 - 9x2 +15x + 3 ( btw it's x cubed and 9x squared ) 1. Find the coordinates of the local and maximum points of the curve y=F(x) to be fair since three people got it right I'll do random number generator online now

 

[golden moon]

LOL, that's pretty easy but you need to use imaginary numbers xD, BTW im not solving the problem cause i don't need the money

 

[N64dude]

x3 - 9x2 + 15x + 3 right or wrong? oh wait i posted the question not the answer.

 

[xanisha]

Sorry but I might just be silly but what are we solving for derivative or for all zeros or x intercepts?
I took the derivative just cause~ F'(x)= 3x^2 -18x+15
Edit I just saw your edit~

 

[xxDianaxx]

x3 - 9x2 + 15x + 3 right or wrong? oh wait i posted the question not the answer.

Yeah was about to say that xD

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Sorry but I might just be silly but what are we solving for derivative or for all zeros or x intercepts?
I took the derivative just cause~ F'(x)= 3x^2 -18x+15

Sorry edited the post ur looking for the local max and min points

 

[N64dude]

xD i have learned algebra before. Its not that hard when you know the basics.

 

[xxDianaxx]

xD i have learned algebra before. Its not that hard when you know the basics.

Yeah

 

[sassystag]

*pulls out graphing calculator*

Local Minimum: x=5
Local Maximum: x=1

?? Not that good at math, is this right? XD

 

[Ril]

Um, minimum is (5,-22) and maximum is (1,10). Is that what you were looking for?

 

[tinytaylor]

were doing this right now in math lmao

 

[Chowder]

Setting f'(x)=0 you get that

x=1 (local max) and x=5 (local min)

Plugging it back to the equation

f(1)= 10 @ x=1
f(5)= -22 @x=5

 

[Shirohibiki]

>not hard once you learn algebra

I've taken like 3 algebra classes and I'm pretty sure you just spoke in moonspeak to me. Holy hell. I felt nauseous just seeing a math problem.

I'M NEVER GOING BACK TO SCHOOL /fLEES

(Anyway this is very generous but math is too scary haha~)

 

[Eleaf]

f(x)=x^3 - 9x^2 +15x + 3
First you find the derivative which is: f'(x)=3x^2-18x+15
Then you use can use then quadratic formula or use that other method, but I'd preferably use the quadratic formula.
-b?√b^2-(4ac)/2a → -(-18)?√(-18)^2-(4*3*15)/2*3 → (18?√324-180)/6 → (18?√324-180)/6 → (18?√144)/6 → (18?12)/6
and then solve by subtracting and adding to get the two x values:
Subtraction: (18-12)/6 → x = 1
Addition: (18+12)/6 → x = 5
And then plug them into the original equation to get y or f(x):
f(x)=x^3 - 9x^2 +15x + 3 → f(1)=(1)^3-9(1)^2+15(1)+3 → f(1) = 10.
f(x)=x^3 - 9x^2 +15x + 3 → f(5)=(5)^3-9(5)^2+15(5)+3 → f(5) = -22.
∴ your coordinates are (1,10) and (5,-22).
(1,10) = Maximum.
(5, -22) = Minimum.

Yay, Calculus. xD

 

[xxDianaxx]

*pulls out graphing calculator*

Local Minimum: x=5
Local Maximum: x=1

?? Not that good at math, is this right? XD

Those point s are right bit ur missing the ys

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Um, minimum is (5,-22) and maximum is (1,10). Is that what you were looking for?

Well done I got right

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f(x)=x^3 - 9x^2 +15x + 3
First you find the derivative which is: f'(x)=3x^2-18x+15
Then you use can use then quadratic formula or use that other method, but I'd preferably use the quadratic formula.
-b?√b^2-(4ac)/2a → -(-18)?√(-18)^2-(4*3*15)/2*3 → (18?√324-180)/6 → (18?√324-180)/6 → (18?√144)/6 → (18?12)/6
and then solve by subtracting and adding to get the two x values:
Subtraction: (18-12)/6 → x = 1
Addition: (18+12)/6 → x = 5
And then plug them into the original equation to get y or f(x):
f(x)=x^3 - 9x^2 +15x + 3 → f(1)=(1)^3-9(1)^2+15(1)+3 → f(1) = 10.
f(x)=x^3 - 9x^2 +15x + 3 → f(5)=(5)^3-9(5)^2+15(5)+3 → f(5) = -22.
∴ your coordinates are (1,10) and (5,-22).
(1,10) = Maximum.
(5, -22) = Minimum.

Yay, Calculus. xD

Ik love calculus u got it also

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Setting f'(x)=0 you get that

x=1 (local max) and x=5 (local min)

Plugging it back to the equation

f(1)= 10 @ x=1
f(5)= -22 @x=5

There also right

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were doing this right now in math lmao

Same as me