application of derivatives (maxima minima)
Asked to find the least value of a€R for which 4ax²+1/x≥1, for all x>0
I@Apu
I@Apu
INote for OP
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BNow in here saying f(x)= 4ax²+1/x, if we calculate f'(1)=0 then that should work right?
BAs f(x)≥1 then 1 must be the point of minima as one of the roots of f'(x)?
SYou could go by (f'=8ax-1/x^2=0 and then find minima)
SOr by AM-GM
BYeah i can put x=1 then and calculate a right as x=1 is the point of minima?l
S$x=\frac{1}{2a^{\frac{1}{3}}}$, then value at minima is $4a(\frac{1}{2a^{\frac{1}{3}}})^{2}+\frac{1}{\frac{1}{2a^{\frac{1}{3}}}}$
TSirLancelotDuLac
SNo x=1 is not minima.
Bwhy not? f(x)≥1:/?
BYeah i that part that's the value of f(1)
SYe the minimum value of f(x) is1. Not the value of x
BOh oh oh oh oh
Bi got it thanks
SNo that is the value of the minima. We put that value >=1 and then get the answer.
B+solved @SirLancelotDuLac
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