Posted by Rochelle F on Wednesday, April 11, 2012 at 8:45pm.
f = x - 1/x + 2/x^3
f' = 1 + 2/x^2 - 6/x^4
= (x^4 + 2x^2 - 6)/x^4
f'=0 when x^4 + 2x^2 - 6 = 0
x^2 = -1 ± √7
-1 - √7 < 0. so we have
x^2 = -1 + √7
x = ±√(√7 - 1)
we want x in (0,∞), so
f(1.28287) = 1.451
or, more exactly,
f(√(√7 - 1)) = √((97√7 - 143)/54)
Related Questions
Pre-Calculus - I need help-I do not understand. The question is this: Write 2x+y...
calculus - PLEASE write 2x+y=5 in polar form. a. -sqrt5 = r cos (theta -...
ALGEBRA 1 - change 50 to 25*2 sqrt 50 = 5 sqrt2 change 8 to 4*2 Nope: (sqrt3-...
algebra - what is the perimeter of a square with sides that are sqrt3 + sqrt5 on...
algebra - evaluate and simplify 6 - square root20 ----------------- 2 The square...
algebra - Can someone please check my work on this problem? For some reason, I ...
Algebra II - I keep getting different answers, can someone help? (16^sqrt5...
math - From the question before, I got to 18+6*SQRT5+6*SQRT5+12 am I on the ...
Calculus (pleas help!!!) - Find the absolute maximum and absolute minimum ...
Calculus (pleas help!!!) - Find the absolute maximum and absolute minimum values...
For Further Reading
