If `f(x)=e^(px+q)` , then show that
`f(a).f(b).f(c)=f(a+b+c).e^(2q)`

If f(x)=e^(px+q) ,then shoq that f(a).f(b).f(c)=f(a+b+c).e^(2q)

If f(x)=e^(px+q) [p and q are constants], show that, f(a).f(b).f(c)=f(a+b+c).e^(2q)

If f(x)=e^(px+2), then f(a)f(b) is

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

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

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

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

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 < de, then the minimum number of zeroes of g(x) = f'(x)^2+f''(x)f(x) in the interval [a, e] is

If f(x)<0,\ x in (a , b) then at the point C(c ,\ f(c)) on y=f(x) for which F(c) is a maximum, f^(prime)(c) is given by a. f^(prime)(c)=(f(b)-f(a))/(b-1) b. \ f^(prime)(c)=(f(b)-f(a))/(a-b) c. f^(prime)(c)=(2(f(b)-f(a)))/(b-a) d. f^(prime)(c)=0