Calc Analysis Homework Help Please?

Question

Note: Yes, this is homework. But no, I am not intending to copy straight the answers from here, however if you can work me through most of it (as opposed to just hints) that would be helpful. This way, I can take your lead, work my way though it, and have help should I get stuck again. Thank you!

8. Suppose that f is differentiable with derivative f '(x) = (1 + x^3)^(-1/2). Show that g = f ^ (-1) satisfies g''(x) = (3/2)*g(x)^2.

22. a. Prove that an incresing and decreasing function intersect at most once.
b. Find two continuous increasing functions f and g such that f(x) = g(x) precisely when x is an integer.

25.b. Prove that if a function is differentiable and nondecreasing, then f'(x) >= 0 for all x.
c. Prove that if f'(x) >= 0 for all x, then f is nondecreasing.

13. Supopse that f is a continuous increasing function with f(0) = 0. Prove that for a,b > 0 we have:
a*b =< integral (0 to a) f(x)dx + integral (0 to b) f ^ (-1) (x)dx
and that equality holds if and only if b = f(a).

Answer 1

#8: We start with something we know: g is the inverse of f. So we have:
g(f(x)) = x
Taking the derivative of both sides (using the chain rule):
d(g(f(x)))/dx = g'(f(x)) * f'(x) = 1
Therefore:
g'(f(x)) = 1/f'(x) = √(1+x³)
Substitute g(x) for x:
g'(f(g(x))) = g'(x) = √(1+g(x)³)
Take the derivative:
g''(x) = 1/(2√(1+g(x)³)) * 3g(x)² * g'(x)
But since g'(x) = √(1+g(x)³), this becomes:
g''(x) = 3/2 g(x)²
As required.

#22a: Let h(x) = f(x)-g(x). Clearly, f(x) intersects g(x) at a point c iff h(c)=0. Since f(x) is increasing, and g(x) is decreasing, it follows that h(x) is increasing. Now suppose that f(x) intersects g(x) at two distinct values of x: c_1 and c_2. Then either c_1>c_2 or c_2>c_1. Since h(x) is increasing, this implies that h(c_1)>h(c_2) or h(c_2)>h(c_1), respectively. Thus in either case h(c_1) ≠ h(c_2). However, since f(c_1)=g(c_1) and f(c_2)=g(c_2), h(c_1) = h(c_2) = 0, contradicting our previous result. So the initial assumption, that f(x) intersects g(x) at two distinct points, is false. Q.E.D.

#22b: Let f(x)=x and g(x) = floor(x) + (x-floor(x))²

#25b: f'(x) = [h→0]lim (f(x+h) - f(x))/h. Suppose that f'(x) < 0. Let ε be less than -f'(x) (this value is greater than zero, since f'(x) is assumed negative). Then by the definition of a limit, ∃δ > 0: ∀h with |h|<δ, (f(x+h) - f(x))/h < 0. In particular, this is true for h such that 0 x, so this contradicts the definition of f as nondecreasing. Therefore, since f'(x) exists, it must be greater than or equal to zero. Q.E.D.

#25c: Let a>b. By the fundamental theorem of calculus, f(a)-f(b) = [b, a]∫f'(x) dx. Since the range of this integral is positive (b
#13: Beats me.

Answer 2

8. I don't know much about this, but I'll have a go at feeling my way.
For y = f(x), then x = g(y).

g'(y) = dx/dy
= 1/(f'(x))
= 1/(f'(g(y)))

and so
g'(x) = 1/(f'(g(x)))

This means, for example, that if
f(x) = x/(sqrt(1 + x^2)), x >= 0
and
g(x) = x/(sqrt(1 - x^2)), 0 <= x < 1,
then
f'(x) = 1/((1 + x^2)^(3/2))

and g'(x) = (1 + (g(x))^2)^(3/2)
= (1 + (x^2)/(1 - x^2))^(3/2)
= 1/((1 - x^2)^(3/2)), which is also obtained by differentiating g(x) directly.

Try differentiating g'(x) = 1/f'(g(x)):

g''(x) = -(1/(f'(g(x))^2)*f''(g(x))*g'(x)
= -(1/f'(g(x))^3)*f''(g(x))

Since f''(x) =-((3x^2)/2)*(1+x^3)^(-3/2), this becomes

g''(x) = -((1+(g(x))^3)^(3/2)*-((3(g(x))^2)/2)*(1+(g(x))^3)^(-3/2)
= 3((g(x))^2)/2
Amazing! I didn't think I was going to solve it! But that's the desired result.

Don't think I can do the others, but might manage if I had time. Sorry, have to go now.

Answer 3

U surely desire to take a trig direction. enable me provide you a wreck down of generic coursework. arithmetic/Pre-Algebra, Algebra, Geometry or Algebra II, Trigonometry, Pre-Calc/greater Algebra you ought to wreck out with taking Pre-Calc and Trig concurrently yet many Pre-Calc classes require you to be attentive to some trig. whether the instructor/professor covers that cloth interior the initiating/middle or end of the direction varies. turn throughout the time of the textual content cloth of the pre-calc direction you would be taking and notice the place the trig comes into play. I taught severe college arithmetic for 11 years.