Question

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 that f''(z)>0.

Answer #1

6. Let a < b and let f : [a, b] → R be continuous. (a) Prove
that if there exists an x0 ∈ [a, b] for which f(x0) 6= 0, then Z b
a |f(x)|dxL > 0. (b) Use (a) to conclude that if Z b a |f(x)|dx
= 0, then f(x) := 0 for all x ∈ [a, b].

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...

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

Let A, B ⊆R be intervals. Let f: A →R and g: B →R be
diﬀerentiable and such that f(A) ⊆ B. Recall that, by the Chain
Rule, the composition g◦f: A →R is diﬀerentiable as well, and the
formula
(g◦f)'(x) = g'(f(x))f'(x)
holds for all x ∈ A. Assume now that both f and g are twice
diﬀerentiable.
(a) Prove that the composition g ◦ f is twice diﬀerentiable as
well, and ﬁnd a formula for the second derivative...

Let 0 < a < b < ∞. Let f : [a, ∞) → R continuous R at
[a, b] and f decreasing on [b, ∞). Prove that f is bounded
above.

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 .

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.

Let D ⊆ R, a ∈ D, let f, g : D −→ R be continuous functions. If
limx→a f(x) = f(a) and limx→a g(x) = g(a) with f(a) < g(a), then
there exists δ > 0 such that x ∈ D, 0 < |x − a| < δ =⇒
f(x) < g(x).

Recall the Mean Value Theorem: If f : [a, b] → R is continuous
on [a, b], and differentiable on (a, b), then there exists c ∈ (a,
b) such that f(b) − f(a) = f 0 (c)(b − a). Show that this is
generally not true for vector-valued functions by showing that for
r(t) = costi + sin tj + tk, there is no c ∈ (0, 2π) such that r(2π)
− r(0) = 2πr 0 (c).

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

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