Calculus Problem Solutions: Temperature Change, Function Derivatives, and Optimal Fencing, Exams of Calculus

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6. 5 pts each Let f be a function such that

lim h→ 0

f(3 + h) − f(3) h

a. Is it possible that

xlim→ 3

f(x) − f(3) x − 3

Justify your answer. No, because limh→ 0 f^ (3+h h) −f^ (3) = limx→ 3 f^ (x x)−−f 3 (3). Both are ex- pressions for f′(3).

b. If f(5) = 12 and f(3) = 2, what is the average rate of change of f on [3,5]? The difference quotient f^ (5) 5 −−f 3 (3) is the expression for the average rate of change of f on [3,5]. This becomes (^125) −− 32 = 5.

7. 3 pts each Suppose that f is continuous at x = 3 and that

lim x→ 3 f(x) = 17.

Indicate whether each statement about f MUST be true, MIGHT be true, or CANNOT be true. Justify your answers.

a. 3 is in the domain of f. MUST be true. If f is continuous at x = 3, f is defined at x = 3 (in other words, 3 is in the domain of f).

b. f(3) = 17. MUST be true. If f is continuous at x = 3, limx→ 3 f(x) = f(3), so f(3) = 17. c.

lim x→ 3 −^

f(x) = 17.

MUST be true. Since f is continuous at x = 3, limx→ 3 f(x) exists, so limx→ 3 −^ f(x) = limx→ 3 +^ f(x) = 17.

8. 4 pts each For each of the functions below, find f′(x).

a. f(x) =

x^3 + 4 f′(x) = 32 x^1 /^2

b. f(x) = 4/

x + π

7 x f′(x) = − 2 x−^3 /^2 + 12 π

7 x−^1 /^2

c. f(x) = 4ex^ − 3 / 5

x^3 f′(x) = 4ex^ + 95 x−^8 /^5

9. 5 pts each A gardener wants to enclose 1000 square meters of land using as little fencing as possible. The garden is to be rectangular; one side, unfenced, lies along a river.

a. Let x be the two sides of the garden perpendicular to the river. Find an expression for L(x) the total length of the fence, in terms of x. L(x) = 2x + (^1000) x. [See Example 2, p. 113)

b. Using calculus, find the minimum amount of fencing needed. Show your work. Find L′(x) and set it equal to zero. L′(x) = 2 − (^1000) x 2 = 0. This means x =

500 ≈ 22 .36 meters. So y = √^1000500 = 2

meters. Need to check that x =

500 is a minimum. Use your graphing calculator or the second derivative test. L′′(x) = (^2000) x 3 so L′′(

500. 0 and L(x) is concave up at x =

500, so it is a local min. Therefore, the minimum amount of fencing needed is: L(x) = 2

500 ≈ 89 .44 meters.

10. A diver’s height above the ground is given by h(t) = t^2 − 7 t + 12 feet on the interval [0, 4] (with t measured in seconds).

a. 2 pts What is the diver’s velocity at time t? v(t) = h′(t) = 2t − 7.