Question
Did I do these two problems right?
P(1-sqrt2) = 8x - 4x^2
P(1-sqrt2)= -4
P(2/x)= 8x - 4x^2
P(2/x) = x-1
I think when the answers are plugged back in, the equation is supposed to equal zero, but I don't get this; however, I always redo each problem with different methods and still get the same answer. Are these the right answers? Any help is GREATLY appreciated! :D
P(1-sqrt2) = 8x - 4x^2
P(1-sqrt2)= -4
P(2/x)= 8x - 4x^2
P(2/x) = x-1
I think when the answers are plugged back in, the equation is supposed to equal zero, but I don't get this; however, I always redo each problem with different methods and still get the same answer. Are these the right answers? Any help is GREATLY appreciated! :D
Answers
Emily
By the way, in the first problem in the parentheses it's supposed to be like "one minus the square root of 2". Sorry if that wasn't clear :-/
Reiny
so the original function is
f(x) = 8x - 4x^2
f(1-√2) = -4 like you had, but ...
f(2/x) = 8x(2/x) - 4(4/x^2)
= 16 - 16/x^2
f(x) = 8x - 4x^2
f(1-√2) = -4 like you had, but ...
f(2/x) = 8x(2/x) - 4(4/x^2)
= 16 - 16/x^2
Anonymous
That made perfect sense the way you had it written out, except I think an extra 'x' was accidentally added in "8x(2/x)". Is it supposed to look like 8(2/x)? Btw, thanks sooooooo much for your help! :D
Emily
Sorry the last "anonymous" was me. xD
Reiny
good for you to catch my error
you are right it should have been
f(2/x) = 8(2/x) - 4(4/x^2)
= 16/x - 16x^2
you are right it should have been
f(2/x) = 8(2/x) - 4(4/x^2)
= 16/x - 16x^2
Emily
Thank you!! :D