If `f : [-5, 5] to R` is a differentiable function function and if f'(x) does not vanish anywhere, then prove that `f(-5) ne f(5)`.

If f:[-5,5]vecR is differentiable function and if f^(prime)(x) does not vanish anywhere, then prove that f(-5)!=f(5)dot

Suppose that f(x) is a differentiable function such that f'(x) is continuous, f'(0)=1 and f''(0) does not exist. Let g(x) = xf'(x). Then

Suppose that f(x) is a differentiable function such that f'(x) is continuous, f'(0)=1 and f''(0) does not exist. Let g(x)=xf'(x) . Then-

Suppose that f(x) is a differentiable function such that f'(x) is continuous, f'(0) = 1 and f"(0) does not exist. Let g(x) = xf'(x). Then

Let f:(0,oo)->R be a differentiable function such that f'(x)=2-f(x)/x for all x in (0,oo) and f(1) =1 , then

If f(0) = 0 ,f'(0) = 2 , then the value differentiable at x = 0 of function y = f[f{f(x)}] is _

Let f(x)=x^(3)-9x^(2)+30x+5 be a differentiable function of x , then -

Prove that the function f(x)=|x| is not differentiable there at x=0.

Let f:Rto R be a twice continuously differentiable function such that f(0)=f(1)=f'(0)=0 . Then

Let f be a differentiable function such that f(1) = 2 and f'(x) = f (x) for all x in R . If h(x)=f(f(x)) , then h'(1) is equal to