Question

Let f : R → R be a bounded differentiable function. Prove that for all ε > 0 there exists c ∈ R such that |f′(c)| < ε.

Answer #1

Let f: R --> R be a differentiable function such that f' is
bounded. Show that f is uniformly continuous.

Let a < b, a, b, ∈ R, and let f : [a, b] → R be continuous
such that f is twice differentiable on (a, b), meaning f is
differentiable on (a, b), and f' is also differentiable on (a, b).
Suppose further that there exists c ∈ (a, b) such that f(a) >
f(c) and f(c) < f(b).
prove that there exists x ∈ (a, b) such that f'(x)=0.
then prove there exists z ∈ (a, b) such...

Let f be a function differentiable on R (all real
numbers). Let y1 and y2 be pair of numbers (y1 < y2) with the
property f(y1) = y2 and f(y2) = y1. Show there exists a num where
the value of f' is -1. Name all theroms that you use and explain
each step.

Let I be an interval. Prove that if f is
differentiable on I and if the derrivative f' be bounded on I then
f uniformly continued on I!

a) Let f : [a, b] −→ R and g : [a, b] −→ R be differentiable.
Then f and g differ by a constant if and only if f ' (x) = g ' (x)
for all x ∈ [a, b].
b) For c > 0, prove that the following equation does not have
two solutions. x3− 3x + c = 0, 0 < x < 1
c) Let f : [a, b] → R be a differentiable function...

Prove or give a counterexample: If f is continuous on R and
differentiable on R∖{0} with limx→0 f′(x) = L, then f is
differentiable on R.

Prove or give a counter example: If f is continuous on R and
differentiable on R ∖ { 0 } with lim x → 0 f ′ ( x ) = L , then f
is differentiable on R .

let H be a subgroup of R. Assume ∃ ε ∈ R ε>0 such that (-ε
,ε) ∩ H= {0}
Prove H cyclic

Let f : E → R be a differentiable function where E = [a,b] or E
= (−∞,∞), show that if f′(x) not = 0 for all x ∈ E then f is
one-to-one, i.e., there does not exist distinct points x1,x2 ∈ E
such that f(x1) = f(x2). Deduce that f(x) = 0 for at most one
x.

Suppose f is differentiable on a bounded interval (a,b) but f is
unbounded there. Prove that f' is also unbounded in (a,b). Is the
converse true?

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