33y2y1 -3y1 +1 =0
y1 = 1/(3-3y2)
y2 = 6yy1/(3-3y2)2
now we know y at -10root(2) .......
we can also get y1 frm the second equation ......
thus we can plug in all values into eq. 3 and get answer!!!!!!!!!
Consider the differentiable function f(x) defined implicitly by the eqn y^3-3y+x=0 in the interval (-\propto ,-2)\bigcup{(2,\propto) }
Given f(-10\sqrt{2})=2\sqrt{2} find f''(-10\sqrt{2})...
A small hint will do....i mean will we have to get y in terms of x...or what? why is the value f(-10√2) given?
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4 Answers
3
3if we actually go on to find f(x) ..... we see its not defined in the region (-2,2) ......
its just un-neccesary ..... i think!!