If f(x) is twice differentiable and f(a)=0,f(b)=2,f(c)= -1,f(d)=2,f(e)=0,where a<b<c<d<e then find the minimum number of zeroes of g(x)={f'(x)2 +{f''(x)f(x)} in [a,e]
nw h(x) = 0 when f(x) = 0 or f`(x) =0 btw [a,e] f(x) = 0 is at two pts i.e x=a and x=e and btw (b,c) and (c,d) f(x) changes sign so btw these intervals f(x) =0 at two more pts thus, f(x) =0 at exactly four pts btw[a,e]
nw f`(x) = 0 at three pts see btw (a,c) there wl be a local maximum at which f`(x) =0 and btw (b,d) there wl be a local minimum at which f`(x) =0 and btw ( c,e) there wl be a local maximum at which f`(x) = 0
so h(x) = 0 at seven points nw applying rolle theorem , h`(x) =0 for atleast six pts
Moderation Team
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