y=logx/x^2 base of log is e find max value of y
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y=f(x)=log(e)x/x²
now,
f`(x)=d/dx{ln(x)/ln(e)x²}
=>f`(x)=(2xln(x)-x)/x⁴
For the existing maximal value of f(x), f`(x) has to be equal to 0.
(2xln(x)-x)/x⁴=0
=>2ln(x)-1=0
=>ln(x)=1/2
=>x=e^(1/2)
Now,
f``(x)={x²-2(2ln(x)-1)x²}/x^6
=>f``(x)=(-3-4ln(x))/x⁴
for x=e^(1/2),
f``(x)=(-3-2)/e²
=>f``(x)=-5/e² [Maximum value exists.]
Now,
for x=√e,
f(x)[max]=log(e)√e/e
=>y[max]=1/2e
now,
f`(x)=d/dx{ln(x)/ln(e)x²}
=>f`(x)=(2xln(x)-x)/x⁴
For the existing maximal value of f(x), f`(x) has to be equal to 0.
(2xln(x)-x)/x⁴=0
=>2ln(x)-1=0
=>ln(x)=1/2
=>x=e^(1/2)
Now,
f``(x)={x²-2(2ln(x)-1)x²}/x^6
=>f``(x)=(-3-4ln(x))/x⁴
for x=e^(1/2),
f``(x)=(-3-2)/e²
=>f``(x)=-5/e² [Maximum value exists.]
Now,
for x=√e,
f(x)[max]=log(e)√e/e
=>y[max]=1/2e
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