12.7. EXERCISES 227

Theorem 12.6.1 Let C be closed and bounded and let f : C→ R be continuous. Then fachieves its maximum and its minimum on C. This means there exist, x1,x2 ∈C such thatfor all x ∈C,

f (x1)≤ f (x)≤ f (x2) .

There is also the long technical theorem about sums and products of continuous func-tions. These theorems are proved later in this chapter.

Theorem 12.6.2 The following assertions are valid.

1. The function af + bg is continuous at x when f , g are continuous at x ∈ D(f)∩D(g) and a,b ∈ R.

2. If and f and g are each real valued functions continuous at x, then f g is continuousat x. If, in addition to this, g(x) ̸= 0, then f/g is continuous at x.

3. If f is continuous at x, f (x) ∈D(g)⊆Rp, and g is continuous at f (x) , then g ◦fis continuous at x.

4. If f = ( f1, · · · , fq) : D(f)→ Rq, then f is continuous if and only if each fk is acontinuous real valued function.

5. The function f : Rp→ R, given by f (x) = |x| is continuous.

12.7 Exercises1. Let f (t) =

(t, t2 +1, t

t+1

)and let g (t) =

(t +1,1, t

t2+1

). Find f ·g.

2. Let f,g be given in the previous problem. Find f ×g.

3. Let f (t) =(t, t2, t3

),g (t) =

(1, t2, t2

), and h(t) = (sin t, t,1). Find the time rate of

change of the box product of the vectors f,g, and h.

4. Let f (t) = (t,sin t). Show f is continuous at every point t.

5. Suppose |f (x)−f (y)| ≤ K |x−y| where K is a constant. Show that f is every-where continuous. Functions satisfying such an inequality are called Lipschitz func-tions.

6. Suppose |f (x)−f (y)| ≤ K |x−y|α where K is a constant and α ∈ (0,1). Showthat f is everywhere continuous. Functions like this are called Holder continuous.

7. Suppose f : R3→ R is given by f (x) = 3x1x2 +2x23. Use Theorem 12.4.2 to verify

that f is continuous. Hint: You should first verify that the function πk : R3 → Rgiven by πk (x) = xk is a continuous function.

8. Show that if f : Rq→ R is a polynomial then it is continuous.

9. State and prove a theorem which involves quotients of functions encountered in theprevious problem.