MthT 430 Chapter 9a Spivak Problem Remarks

7.

f(x) = x³. f^′(x) = 3 x²

(a)

f^′(9) = 3 ·9 = 27.

(b)

f^′(3²) = f^′(9) = 27 or f^′(3²) = 3 ·(3²)² = 27.

(c)

f^′(a²) = 3·(a²)² = 3 a⁴; f^′(x²) = 3·(x²)² = 3x⁴.

(d)

f(x) = x³; f^′(x) = 3 x²; f^′x² = 3 x⁴. g(x) = f(x²) = x⁶; g^′(x) = 6 x⁵.

8.

(a)

g(x) = f(x+c).

g′(x) = lim h → 0  g(x + h) − g(x) h = lim h→ 0  f(x + c + h) − f(x + c) h = f′(x + c).

(b)

g(x) = f(cx). For c ≠ 0,

g′(x) = lim h → 0  g(x + h) − g(x) h = lim h → 0  f(c(x + h)) − f(c x) h = lim ch → 0  c · f(cx + ch)0 − f(c x) ch =c ·f′(cx).

10.

f(x) = g(t +x). f^′(a) = g^′(t + a); f^′(x) = g^′(t + x ).

F(t) = g(t +x). F^′(a) = g^′(a + x); F^′(x) = g^′(x + x ).

11.

(a)

If s′ is proportional to s, there is a constant k such that s′(t) = k s(t). For s(t) ≠ 0, s′(t)/s(t) is constant.

If S(t) = c t², S′(t) = 2 c t. For t ≠ 0, S(t) ≠ 0, and S′(t)/S(t) = 2/t, which is not constant.

(b)

If s(t) = (a/2) t²,

s′(t) = a t. s′′(t) = a.

Note that

(s′(t))2 = (a t)2 = 2 a s(t).

12.

Speed limit at position x is L(x). Position of A at time t is denoted by a(t).

(a)

A travels at the speed limit means: For all t, a′(t) = L(a(t)).

(b)

Suppose A travels at the speed limit and b(t) = a(t −1). Then b′(t) = a′(t−1) = L(a(t − 1)) = L(b(t)), and B travels at the speed limit.

(c)

If b(t) = a(t) − k, b′(t) = a′(t) = L(a(t)). Then b′(t) = L(b(t)) for all t, if and only if L(b(t)) = L(a(t) − k) = L(a(t)), or L(x) is periodic with period k.

18.

f is the oneoverq function. If r is a rational number, f is not continuous at r. Thus f is not differentiable at r.

If a is an irrational number, f(a) = 0. If h is rational, the difference quotient is 0. Thus if f′(a) exists, f^′(a) = 0.

Let a have the nonrepeating decimal expansion m.a₁ a₂ …a_n …. Define the irrational number h_n = −0.00…0 a_n a_{n+ 1}…, so that a + h_n = m.a₁ a₂… a_n−1.

Now |h_n| ≤ 10¹⁻ⁿ, |1/h_n| ≥ 10ⁿ⁻¹, and f(a + h_n) = 1/q with q ≤ 10ⁿ⁻¹ so that |f(a + h_n)| = 1/q ≥ 10¹⁻ⁿ.

It follows that

⎢ ⎢ f(a+hn)− f(a) hn ⎢ ⎢ = |f(a +hn)| |hn| = 1/q |hn| ≥ 101 −n 10n−1 = 1.

Conclude that f^′(a) does not exist.

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