1. Let f(x) and g(x) be functions defined on a common domain. Suppose f(x) is co

Question

1. Let f(x) and g(x) be functions defined on a common domain. Suppose f(x) is continuous at x0 and (f+g)(x) is continuous at x0. Show that g(x) is continuous at x0 using a definition of continuity at a point.

2. f(0) = f(1); yet its derivative is never zero on (0,1). Explain why this does not contradict Rolle's Theorem.

3 Suppose g(x) is required to be differentiable on the interval ( - 10, 10). Can g then have the following characteristics: g (-2) = - 2; g (2) = 6; g ' (x) < 1 for all x in ( -10, 10)? Completely justify your answer.

Explanation / Answer

1)ans.To prove f(x) is continuous at x0 for any e > 0 I need to find d so that |x - x0| < d => |f(x) - f(x0)| < e

So first assume that x0 >= 0. Since f(x) = f(-x) proving continuity for x0 >= 0 suffices.
| x^2 - x0^2| <= |(x0 + d)^2 - x0^2| = |2x0d + d^2| = 2x0d + d^2

So for any e > 0 I will choose d so that e > 2x0d + d^2. It's clear that I can do that because the right side can be made arbitrarily small for some d.

Note that the x0 on the right means that x^2 is not uniformly continuous, just continuous

2)ans.

If f is didifferentiable on the open

interval (a; b), continuous on the closed interval [a; b] and f (a) = f (b) then

there exists a c : a < c < b such that f0 (c) = 0

Since f is continuous on the closed interval [a; b] it is bounded and its

bounds are attained. So there exist k; ` 2 [a; b] such that f (k) f (x) f (`)

for all x 2 [a; b].

If f (k) = f (`) then f (x) equals this common value for all x thus f is

constant. Hence f0 (x) = 0 for all x 2 [a; b]. Choose c = (a + b) =2.

Otherwise f (k) < f (`). Thus at least one of f (k) and f (`) diers from

the common value of f (a) = f (b) : Assume f (k) diers from f (a) and

f (b). Firstly, f (k) 6= f (a) means that k 6= a while f (k) 6= f (b) implies

k 6= b. Thus k 2 (a; b), which means that f is dierentiable at k. Then

f (k) f (x) for all x 2 (a; b) means that k is a local minimum of f and

thus, by the Theorem above, satises f0 (k) = 0.

I leave it to the interested student to show that if f (l) diers from f (a)

and f (b) then f0 (l) = 0.

so f(0) = f(1); yet its derivative is never zero on (0,1) in Rolle's Theorem.

3)ans.

to find the formula of f, use the equation: y-y1=m(x-x1)

g'(x)=2x g'(-2)=-6 g is increasing at x=-2 because g'(-2)<1

h'(x)=f'(x)g(x)+g'(x)f(x)

Plug in the values into the equation and you get to be proved and justified the answer

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