Question

Let g from R to R is a differentiable function, g(0)=1, g’(x)>=g(x) for all x>0 and g’(x)=<g(x) for all x<0. Proof that g(x)>=exp(x) for all x belong to R.

Answer #1

Let f: R -> R and g: R -> R be differentiable, with g(x) ≠
0 for all x. Assume that g(x) f'(x) = f(x) g'(x) for all x. Show
that there is a real number c such that f(x) = cg(x) for all x.
(Hint: Look at f/g.)
Let g: [0, ∞) -> R, with g(x) = x2 for all x ≥ 0. Let L be
the line tangent to the graph of g that passes through the point...

4a). Let g be continuous at x = 0. Show that f(x) = xg(x) is
differentiable at x = 0 and f'(0) = g(0).
4b). Let f : (a,b) to R and p in (a,b). You may assume that f is
differentiable on (a,b) and f ' is continuous at p. Show that f'(p)
> 0 then there is delta > 0, such that f is strictly
increasing on D(p,delta). Conclude that on D(p,delta) the function
f has a differentiable...

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

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 and g : [0, 1] → R be continuous, and assume f(x) = g(x)
for all x < 1. Does this imply that f(1) = g(1)? Provide a proof
or a counterexample.

If f is a differentiable function such that f′(x) = (x^2−
16)*g(x), where g(x)>0 for all x, at which value(s) of x does f
have a local maximum?
1. At both x=-4,4
2. Only at x=-16
3. Only at x=4
4. At both x=-16,16
5. Only at x=-4
6. Only at x=16

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.

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

Consider the function g(x) = |3x + 4|.
(a) Is the function differentiable at x = 10? Find out using
ARCs. If it is not differentiable there, you do not have to do
anything else. If it is differentiable, write down the equation of
the tangent line thru (10, g(10)).
(b) Graph the function. Can you spot a point “a” such that the
tangent line through (a, f(a)) does not exist? If yes, show using
ARCS that g(x) is not...

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