The following are problems to try as review.
1. For the following problems,
let P(x) = 2x^7 + 4x^2 - 5, and
let Q(x) = 5x^7 - 18x^4 + x^3 + 11.
(i) lim_(x --> oo) P(X) = ?
Since P(x) is continuous,
(ii) lim_(x --> 0) P(X) = P(0) = -5 (the constant term)
(iii) lim_(x --> 1) P(X) = P(1) = 2 + 4 - 5 = 1 (sum of coefficients)
(iv) lim_(x --> 0) P(X)/Q(x) = -5/11
(v) lim_(x --> oo) P(X)/Q(x) = 2/5 (ratio of leading coefficients
since the two polynomials are of the same degree; if P had
degree greater than that of Q, the limiting ratio would be
+oo times the ratio of the leading coefficients - could you
prove that?)
(vi) lim_(x --> 1) P(X) + 4Q(x) = P(1) + 4Q(1) = 1 + 4*(-1) = -3.
Let R be the function with
R(x) = P(x) + 4sin[(pi/2)x].
(vii) R(1) = P(1) + 4sin((\pi/2)) = -3 + 4*1 = 1
(viii)* (* means harder)
lim_(x --> oo) R(X)/Q(x) = ?
(ix)* lim_(x --> 1) R(X)/Q(x) = ?
(x)* Is it true that P(x) has a root in (0,1)? Y (IVT)
(xi)* Is it true that P(x_0) = 100 for some x_0 in (0,1)? N
(P'(x) > 0 for x>0 so P(x) is increasing on (0,\infy;
hence, P(1) is the largest value of P(x) for x in (0,1).
(xii)* Is it true that P(x_0) = 100 for some x_0 in (1,2)?
Y (IVT)
2. For the following, find the derivative:
(i) sqrt(2x + 5) = (2x + 5)^(1/2) has derivative
(1/2)2 (2x + 5)^(-1/2)
(ii) 1/(4x + 5)
(iii) If lim_(x --> x_0) f(x) = L and
lim_(x --> x_0) g(x) = M,
then
lim_(x --> x_0) [2f(x) + g(x)/f(x) = ?
provided that L and M satisfy ?
(iv) State two versions of the definition of derivative.
(v) What does it mean to say lim_(x --> oo) f(x) = 0?
For each of the next three problems, indicate which of
the pair implies the other and give an example to show
the implication is only one-way:
(vi) {f continuous at x_0, f differentiable at x_0}
(vii) {f cont at x_0 in (a,b), f continuous on (a,b)}
(viii) {lim_(x --> a+) f(x) = lim_(x --> a-) f(x),
lim_(x --> a) f(x) does not exist}
(ix)* Assume that f'(a) exists. Find g'(a) for the function
g(x) = f(x)/x by evaluating it as a limit. Your answer
will be in terms of f'(a), f(a), and a.
(x)* Find lim_(x --> 4+) (4 - x) / |4 - x| (i.e., of
difference divided by absolute value of difference).
(xi)* Find f and a such that (N.B. 64 = 2^6)
f'(a) = lim_(h --> 0)(1/h)( (2 + h)^6 - 64 )
Let D denote the differentiation operator
D(f) = f'
(xii)* For f,g in Dom(D) and a,b real numbers
D(af + bg) = ?
(xiii)* For f,g,h in Dom(D),
D(fgh) = ? where fgh(x) = f(x)g(x)h(x).
(xiv)* For g(x) = f(2x), find g'(x) in terms of f'.
We did this in class.
g'(x) = lim_(x --> a) (f(2x) - f(2a))/(x-a)
= 2 lim_(x --> a) (f(2x) - f(2a))/(2x - 2a)
= 2 lim_(2x --> 2a) (f(2x) - f(2a))/(2x - 2a)
= 2 f'(2x)
(This is a special case of the chain rule which will be proved
later.)
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