anonymous
  • anonymous
What is (x^2+1) times (2x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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UnkleRhaukus
  • UnkleRhaukus
distribute (a+b) x (c) = (a x c) + (b x c)
anonymous
  • anonymous
So (x^2 (times) 2x) + (1 (times) 2x)
UnkleRhaukus
  • UnkleRhaukus
That's right

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More answers

anonymous
  • anonymous
Well I don't know what the first half would be. I don't know how to do the x's So it's something + 2x
UnkleRhaukus
  • UnkleRhaukus
ok well your almost there
UnkleRhaukus
  • UnkleRhaukus
\[An^p\times Bn^q=(A\times B)\times n^{p+q} \]
anonymous
  • anonymous
2x^3?
UnkleRhaukus
  • UnkleRhaukus
YES!
anonymous
  • anonymous
Well that's what I thought it was but it's not one of my choices...
UnkleRhaukus
  • UnkleRhaukus
what was your final answer?
anonymous
  • anonymous
2x^3 + 2x
UnkleRhaukus
  • UnkleRhaukus
Hmm, that is right, what are the choices?
anonymous
  • anonymous
2x^2 +2 2x^2+1 x^2+4x+4 4x^2+1
UnkleRhaukus
  • UnkleRhaukus
Are you sure those are the right options for this question?
anonymous
  • anonymous
Yepp. Here I'll put the whole question here so you can see if I'm trying to do this wrong.
anonymous
  • anonymous
1 Attachment
UnkleRhaukus
  • UnkleRhaukus
ah i see the propbem
UnkleRhaukus
  • UnkleRhaukus
you have found the product of the functions \[(f\cdot g)(x)=f(x)\times g(x)\] the question is looking for the function of a fcunction \[[f\circ g](x)=f\big(g(x)\big)\]
UnkleRhaukus
  • UnkleRhaukus
*function of a function.
UnkleRhaukus
  • UnkleRhaukus
\[f(x)=x^2+1\\g(x)=2x\\ \\ [f\circ g](x)=f\big(g(x)\big)=f(2x)=(2x)^2+1=\]
UnkleRhaukus
  • UnkleRhaukus
\[[g\circ g](x)=g\big(g(x)\big)=g(2x)=2(2x)=4x\]
UnkleRhaukus
  • UnkleRhaukus
have you seen this notation before?
UnkleRhaukus
  • UnkleRhaukus
@RoseDryer
UnkleRhaukus
  • UnkleRhaukus
are you still here?
anonymous
  • anonymous
I'm sorry I was making a bottle.
anonymous
  • anonymous
I have not seen that before
UnkleRhaukus
  • UnkleRhaukus
does it make any sense ?
anonymous
  • anonymous
So it would be f(g(x))
UnkleRhaukus
  • UnkleRhaukus
yeah
anonymous
  • anonymous
So (2x)^2+1
UnkleRhaukus
  • UnkleRhaukus
yes, last step is to simplify that
anonymous
  • anonymous
How do I?
UnkleRhaukus
  • UnkleRhaukus
\[(2x)^2=(2x)\times(2x)=(2\times2)(x\times x)=\]
anonymous
  • anonymous
So 4x^2?
UnkleRhaukus
  • UnkleRhaukus
yep
UnkleRhaukus
  • UnkleRhaukus
so your final answer now is ____ _ _
UnkleRhaukus
  • UnkleRhaukus
yep
anonymous
  • anonymous
4x^2+1
UnkleRhaukus
  • UnkleRhaukus
Correct \[\boxed{\large\color{red}\checkmark}%unk\]
UnkleRhaukus
  • UnkleRhaukus
do you want another simple example with this new notation?
anonymous
  • anonymous
I just want to know why we simplified.
UnkleRhaukus
  • UnkleRhaukus
do you when we simplified (2x)^2 ?
anonymous
  • anonymous
What?
UnkleRhaukus
  • UnkleRhaukus
do you mean* when we simplified (2x)^2 ?
anonymous
  • anonymous
Yeah. Why did we have to do that?
UnkleRhaukus
  • UnkleRhaukus
we did that because our final answer is a quadratic, and a nice general for of a quadratic is ax^2+bx+c=0 where a, b, c are the coefficeints, we simplified to get the coefficent next to the x^2 term
anonymous
  • anonymous
Ahh. Okay
UnkleRhaukus
  • UnkleRhaukus
if (2x)^2+1 was an option that would be correct as well.
anonymous
  • anonymous
It is.
UnkleRhaukus
  • UnkleRhaukus
\(2x^2\) is different to \((2x)^2\)
anonymous
  • anonymous
Oh okay.

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