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Honors Calculus Practice Final Below is a practice final. The actual final will be somewhat shorter. This final is probably about 2.5 hours long rather than just two hours long.
(1) Define the derivative f ′(a). Calculate (from the definition) the derivative of f (x) = x^2.
f ′(a) = lim h→ 0
f (a + h) − f (a) h
If we take f (x) = x^2 , then
f ′(x) = lim h→ 0
(x + h)^2 − x^2 h
= lim h→ 0
2 xh + h^2 h
= lim h→ 0 (2x + h) = 2x.
(2) Let
f (x) =
x^3 x ∈ Q 0 else. Calculate f ′(0) from the definition. First note that f^ ( hh )is either 0 or h^2. So we have
0 ≤ f ′(0) = lim h→ 0
f (h) − f (0) h
≤ lim h→ 0 h^2 = 0.
Therefore, f ′(0) = 0. (3) Suppose f : [0, 1] → [0, 1] is a continuous function defined on the closed interval [0, 1]. Prove f (x) = x for some x ∈ [0, 1]. Consider g(x) = f (x) − x. g(0) ≥ 0 , g(1) ≤ 0. If both inequalities are strict, then the intermediate value theorem tells us that g(x) = 0 for some x ∈ (0, 1), so f (x) = x. If either inequality is an equality, that gives an x where f (x) = x. (4) Prove from the definition of the derivative that if f (x) = |x|, then f ′(0) is not defined. (You’ll need to show a certain limit doesn’t exist.) If the limit defining f ′(0) was a non-negative number l, then for =. 5 , for any δ take h = −δ/ 2. Then |h − 0 | < δ, but ||h|− h 0 − l| ≥ 1. So the limit can’t be l. If the limit defining f ′(0) was a negative number l, then for =. 5 , for any δ take h = δ/ 2. Then |h − 0 | < δ, but ||h|− h 0 − l| ≥ 1. So this limit can’t be l. (5) Prove that
10 exists by showing the function x^2 − 10 has a root. Consider the function f (x) = x^2 − 10. f (0) == 10. f (4) = 6. So by the IVT, f (x) = 0 for some x ∈ (0, 4). This x is
(6) (a) State the least upper bound axiom. Any non-empty set of real numbers which is bounded above has a least upper bound. (b) Prove that if a 1 ≤ a 2 ≤ · · · ≤ an ≤ an+1 ≤ · · · ≤ bn+1 ≤ bn ≤ · · · ≤ b 2 ≤ b 1
then there is some x in all of the [ai, bi]. Since b 1 is an upper bound for the set of ai, the set of ai has a least upper bound. Call this α. Since a 1 is a lower bound for the set of bi, the set of bi has a greatest lower bound. Call this β. 1
Claim: α ≤ β. To see this suppose α > β. Let c be the number midway between α and β. c < α so c is not an upper bound for the set of ai. So there is some an > c. Similarly c > β so c is not a lower bound for the bi. So there is some bm < c. But then bm < an, and this is a contradiction since all the ais are smaller than all the bis. So if x ∈ [α, β] then x ≥ α ≥ an for all n, and x ≤ β ≤ bn for all n. So x is in all [an, bn]. (7) Use induction and the product rule to prove that if f (x) = (^) x^1 n then f ′(a) = (^) a−n+1n. We start with n = 1. Let fn(x) = 1/xn. Then
f 1 ′(a) = lim h→ 0
1 /(a + h) − 1 /a h
= lim h→ 0
a − (a + h) ha(a + h)
= − 1 /a^2.
That was the base case. Now assume that we know that f (^) n′− 1 (a) = −n+ an^. Then since fn(x) = f 1 (x)fn− 1 (x), we can use the product rule to see
f (^) n′(a) =
a^2
fn− 1 (a) + f 1 (a)
−n + 1 an^
a^2
an−^1
a
−n + 1 an^
−n an+ which is the desired result. (8) Find the following limits (keep in mind that some of them may be un- defined). (a) lim x→∞
x x sin x + 7
= lim x→∞
sin x + 7/x This is undefined because the 7/x term goes to 0, and the sin x term varies between −1 and 1. (b)
lim x→∞
x^2 + 4x−
x^2 + x = lim x→∞
(x^2 + 4x) − (xx^ + x) √ x^2 + 4x +
xx^ + x
= lim x→∞
1 + 4/x +
1 + 1/x
(c)
lim x→∞
x^3 + 390x^2 + 10, 000 x^4 + 1
= lim x→∞
1 /x + 390/x^2 + 10, 000 /x^4 1 + 1/x^4
(d)
lim x→−∞
sin(x^2 ) x^2
since the numerator is bounded by ±1 and the denominator goes to ∞. (9) For the following questions, indicate whether each limit is ∞, −∞ or just fails to exist. (a) limx→ (^0 1) x. Fails to exist since x can approach 0 from above or below. (b) limx→ (^0) x^12. Positive infinity since the denominator approaches 0 from above. (c) limx→ (^1) −(1^1 −x) 2. −∞ since the denominator approaches 0 from below.
(13) Graph x − [x], and graph its derivative. Your graph of the derivative should indicate any places where the derivative is undefined.
(14) Take
f (x) =
x^2 x ≥ 0 −x^2 x < 0. Calculate f ′(x). What is f ′(0)? f ′(x) = 2x for x > 0 , f ′(x) = − 2 x for x < 0. At 0 ,
f ′(0) = lim h→ 0
±h^2 − 0 h
= lim h→ 0
±h = 0.
(15) Give two functions f, g so that
f ′(x) = g′(x) =
1 x > 0 − 1 x < 0 and so that f − g is not a constant function. Take f (x) = |x|. Take
g(x) =
x x ≥ 0 −x + 1 x < 0
(16) Calculate some derivatives: (a) (x + 7)^100 100(x + 7)^99. (b) x x−+1^1 Using the quotient rule 1(x + 1) − (x − 1) (x + 1)^2
(x + 1)^2
(c) (^) 1+^1 x Using the quotient rule again, − 1 · (− 1 /x^2 ) (1 + 1/x)^2
x^2 + 2x + 1
(d) (x + 7)^100 (x − 5)^50 100(x + 7)^99 (x − 5)^50 + (x + 7)^100 50(x − 5)^49.
(17) Find the tangent line to f (x) = x^3 at x = 2. Find the points where this line intersects the graph of y = x^3. This is the line through (2, 8) with slope f ′(2) = 12. So it is the line y = 12(x − 2) + 8. This intersects y = x^3 when x^3 = 12(x − 2) + 8. This happens when x^3 − 12 x + 16 = 0. This has the solution x = 2. If we factor out x − 2 , we get x^2 + 2x − 8 = 0. This quadratic factors as (x + 4)(x − 2) = 0. So the line intersects the graph of x^3 at (− 4 , −64) and of course at (2, 8). (18) Answer true or false for each of the below. Supply a short justification if possible. (a) If (f + g)′(a) exists, then f ′(a) and g′(a) exist. FALSE: Take f (x) = |x|, g(x) = −|x|, a = 0. (b) If f is continuous at a then f is differentiable at a. FALSE: Take f (x) = |x|, a = 0. (c) If f is differentiable at a, then f is continuous at a. TRUE: This is a theorem from chapter 9. (d) If f and g are differentiable at a and g(a) 6 = 0, then f /g is differen- tiable at a. TRUE: by the quotient rule. (e) If f is continuous on (a, b), f has an upper bound on (a, b). FALSE: Take f (x) = 1/x, (a, b) = (0, 1). (f) If a = b then f (a) = f (b). TRUE: This is the definition of a function. (g) If f (a) = f (b) then a = b. FALSE: Take f (x) = x^2 , a = 3, b = − 3. (h) If [f (x)]^2 is continuous, then f (x) is continuous. FALSE: Take f (x) =
x x ∈ Q −x else (i) If limh→ 0 f (h) = limh→ 0 g(h) then limh→ 0 (f /g)(h) = 1. FALSE: Take f (h) = h^2 , g(h) = h. Both limits are 0 , but the limit of the quotient is also 0. (j) If f is a polynomial of even degree, f has either a maximum or a minimum value. TRUE: Suppose the coefficient of the highest degree term is positive. Let c be the constant term. Then there is an M so that if |x| > M , f (x) > 2 |c|. Then the minimum value of f (x) on [−M, M ] is the minimum value of f (x) on all of R.
