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determine the values of x given for which f'(x)>0, given that f(x)= 1/(4+2x^2)^4
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that would be easier to do if you simply take and negate the exponent and work the reciprocal
f(x) = (4 +2x^2)^(-4) f'(x) = -4(4 +2x^2)^(-5) 4x
so when x = 0; is all I see for a zero
-16x ----------- = 0 when x = 0 (4 +2x^2)^5
Real solutions anyway.
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yes :) assumably
when x = 1 we get a - value; so stick with values of x<0
thanks!
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