Question

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 (0, -6). Find the point (x0, x02) that is the point of tangency.

Answer #1

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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...

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...

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.

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(x) = x^3 - x
a) Find the equation of the secant line through (0,f(0)) and
(2,f(2))
b) State the Mean-Value Theorem and show that there is only one
number c in the interval that satisfies the conclusion of the
Mean-Value Theorem for the secant line in part a
c) Find the equation of the tangent line to the graph of f at point
(c,f(c)).
d) Graph the secant line in part (a) and the tangent line in part...

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(x)=22−x2f(x)=22-x2
The slope of the tangent line to the graph of f(x) at the point
(−4,6) is .
The equation of the tangent line to the graph of f(x) at (-4,6) is
y=mx+b for
m=
and
b=
Hint: the slope is given by the derivative at x=−4

Problem: Let y=f(x)be a differentiable function
and let P(x0,y0)be a point that is not on the graph of function.
Find a point Q on the graph of the function which is at a
minimum distance from P.
Complete the following steps. Let Q(x,y)be a point on the graph
of the function
Let D be the square of the distance PQ¯. Find an expression for
D, in terms of x.
Differentiate D with respect to x and show that
f′(x)=−x−x0f(x)−y0
The...

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.

1. Let f be the function defined by f(x) = x
2 on the positive real numbers. Find the
equation of the line tangent to the graph of f at the point (3,
9).
2. Graph the reflection of the graph of f and the line tangent to
the graph of f at the point
(3, 9) about the line y = x.
I really need help on number 2!!!! It's urgent!

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