f(x)=2x-4ln(x)
1st derivative=2-4x^-1
2nd derivative=4x^-2
x=2 is the only critical point...
QUESTION STARTS HERE: is x=2 a relative maximum or relative minimum for f? Explain whether you are using the 1st or 2nd derivative.
1st derivative=2-4x^-1
2nd derivative=4x^-2
x=2 is the only critical point...
QUESTION STARTS HERE: is x=2 a relative maximum or relative minimum for f? Explain whether you are using the 1st or 2nd derivative.
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f(x)=2x-4ln(x)
f '(x) =2-4x^-1 = 0
2 = 4/x
x = 2 this is a relative min or a relative max
f"(2) =4 (2)^-2 =1 which is positive. which means x = 2 is a relative minimum.
f '(x) =2-4x^-1 = 0
2 = 4/x
x = 2 this is a relative min or a relative max
f"(2) =4 (2)^-2 =1 which is positive. which means x = 2 is a relative minimum.