f(x)=2x^3 - x^2 + 1/2x Let g(x)=f(2/3x) What is g(x) in terms of x? Please help me out I'm reviewing for eoc and would like to know how to do this
@ganeshie8
\(\LARGE f\left(\frac{2}{3}x\right)\) means you replace every x in f(x) with (2/3)x like so... \[\Large f(x) = 2x^3 - x^2 + \frac{1}{2}x\] \[\Large f({\color{red}{x}}) = 2{\color{red}{x}}^3 - {\color{red}{x}}^2 + \frac{1}{2}{\color{red}{x}}\] \[\Large f\left({\color{red}{{\frac{2}{3}x}}}\right) = 2\left({\color{red}{{\frac{2}{3}x}}}\right)^3 - \left({\color{red}{{\frac{2}{3}x}}}\right)^2 + \frac{1}{2}\left({\color{red}{{\frac{2}{3}x}}}\right)\] Do you see what to do next?
On 2(2/3x)^3 is where I am having the most problem
when you cube something, you're multiplying it by itself 3 times examples 8^3 = 8*8*8 = 512 5^3 = 5*5*5 = 125 7^3 = 7*7*7 = 343 make sense?
Yes, So would it be 6/9x then multiplied by 2? Because if i do 2/3 I get decimals that dont necessarily make sense.
so you'll use that same idea for (2/3x)^3
I have the answer sinceI am using one of the eoc reviews but I am confused as to how they got what they did
\[\Large \left(\frac{2}{3}x\right)^3 = \left(\frac{2}{3}x\right)*\left(\frac{2}{3}x\right)*\left(\frac{2}{3}x\right)\]
Oh okay that makes sense now
\[\Large \left(\frac{2}{3}x\right)^3 = \left(\frac{2}{3}x\right)*\left(\frac{2}{3}x\right)*\left(\frac{2}{3}x\right)\] \[\Large \left(\frac{2}{3}x\right)^3 = \left(\frac{2}{3}*\frac{2}{3}*\frac{2}{3}\right)*(x*x*x)\] \[\Large \left(\frac{2}{3}x\right)^3 = \frac{2*2*2}{3*3*3}*x^3\] \[\Large \left(\frac{2}{3}x\right)^3 = \frac{8}{27}x^3\]
Then multiply the 8 by 2?
8 by the 2? what do you mean?
oh I see now, yes
the 2 from 2x^3
since it was originally 2(2/3x)^3
Yes
`Then multiply the 8 by 2?` yes to get \[\Large 2*\left(\frac{2}{3}x\right)^3 = 2*\frac{8}{27}*x^3=\frac{2*8}{27}*x^3=\frac{16}{27}x^3\]
Gotcha.
so then for (2/3x) ^2 would you basically just do 2*2 and then 3*3 to get 4/9 and x*x so it would be 4/9x^2
very good
how about the last part with 1/2 times (2/3x)
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