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Find the smallest value of A such that the function: t^4-17t^2+16 is increasing for all t in the interval(A,0)?
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- MathsorcererLv 71 decade agoFavorite Answer
A function is increasing when f'(x) > 0.
f'(t) = 4t^3 - 34t = 2t(2t^2 - 17), which will have roots at
t = 0 and t = +/- sqrt(17/2) -->
f'(t) = (t + sqrt(17/2))(2t)(t - sqrt(17/2))
When -sqrt(17/2) < t < 0, the (t + sqrt(17/2)) term is positive and both the (2t) and (t - sqrt(17/2)) terms are negative, so f'(t) will be positive.
A = -sqrt(17/2)
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