SOLUTION: I've gotten this problem wrong twice so I'm obviously doing something wrong.
find f(x/2) when f(x)=x^3-4x^2
my solution: (x/2)^3-4(x/2)^2= (x^3/2^3)-4(x^2/2^2)= x^3/8 - 4x^3/4
Algebra ->
Rational-functions
-> SOLUTION: I've gotten this problem wrong twice so I'm obviously doing something wrong.
find f(x/2) when f(x)=x^3-4x^2
my solution: (x/2)^3-4(x/2)^2= (x^3/2^3)-4(x^2/2^2)= x^3/8 - 4x^3/4
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Question 58148: I've gotten this problem wrong twice so I'm obviously doing something wrong.
find f(x/2) when f(x)=x^3-4x^2
my solution: (x/2)^3-4(x/2)^2= (x^3/2^3)-4(x^2/2^2)= x^3/8 - 4x^3/4
x^3/8 - (4x^3/4)(2)= x^3/8 - 8x^3/8= -7x^3/8
sooo...it looks right to me, but it's not. Any example would be great! Thanks! Answer by hayek(51) (Show Source):