Asked by Emily
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
Answered by
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 :-/
Answered by
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
Answered by
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
Answered by
Emily
Sorry the last "anonymous" was me. xD
Answered by
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
Answered by
Emily
Thank you!! :D
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