Question

If f(x) is a twice differentiable function such that f(a)=0,f(b)=2,f(c)=1,f(d)=2,f(e)=0, where a<b<c<d<e, then the minimum number of zeroes of g(x)=(f′(x))^2+f"(x)f(x) in the interval [a,e] is

Answer 1

f(x) becime zero at 'a' and 'e'. Between f(b)=2 and f(c)=-1 there is a value for x for which f(x)=0 as the graph passes from 2 to -1 it touces the x axix.lly whe it pass from -1 to 2 it there is a value for x such tha f(x)=0.