MATH 1210 - Section 002
Fall 2010
Nicolas Fourrier
Due on September 14, 2010
Homework 3 Name:
1. Explain what it means to say that
lim f (x) = 3 and
x→−1−
lim f (x) = 7.
x→−1+
In this situation is it possible that lim f (x) exists ? Explain. x→−1
As x approaches 1 from the left, f (x) approaches 3; and as x approaches 1 from the right, f (x) approaches 7. No, the limit does not exist because the left- and right- hand limits are different.
−x + 1 2x + 3
2. Let f (x) =
if x ≤ 0 if x > 0
Find lim f (x) and lim f (x). x→0−
x→0+
Is this function continuous everywhere ? lim f (x) = 1 and lim f (x) = 3. No, this function is not continuous at x = 0.
x→0−
x→0+
3. Find the following limits (a) (b) (c) (d)
lim
x→2−
x−3 x+2
lim
x→−2−
x−3 x+2
lim 2x +
√
x→−3+
lim
x→0−
(a) (b) (c) (d)
3 x
lim
x→2−
x−3 x+2
= − 14
x−3 x+2
lim
x→−2−
does not exist.
lim 2x +
x→−3+
lim
x→0−
( 4. Let f (x) =
x+3
3 x
√
x + 3 = −6
does not exit.
x2 −4 x+2
k
if x 6= −2 if x = −2
Find the values of k that will make f continuous on (−∞, ∞). k = −4 because lim f (x) = −4 x→−2
5. Show that the function is continuous for all values of x in the interval [a, b]. Prove that f must have at least one zero in the interval (a, b). Use Intermediate Value Theorem (a) f (x) = 2x2 − 3x2 − 36x + 14; a = 0, b = 1 (b) f (x) = 2x5/3 − 5x4/3 ; a = 14, b = 16
For each function, use the properties of continuous functions. Then compute f (a) and f (b), observe they have opposite signs and use the modified version of the intermediate value theorem. 6. Suppose f is continuous everywhere except at x = 2, and f (1) = 0, f (4) = 7. Does there necessarly exist at least one number c in [1, 4] such that f (c) = 3?
x − 1 if x ≤ 2 x + 3 if x > 2 Then f is continuous everywhere except at x = 2, and f (1) = 0, f (4) = 6, but there is no number c in [1, 4] such that f (c) = 3 Consider the following function: f (x) =
7. A Tibetan monk leaves the monastery at 7:00 A.M. and takes his usual path to the top of the mountain, arriving at 7:00 P.M.. Then following morning, he starts at 7:00 A.M. at the top and takes the same path back, arriving at the monastery at 7:00 P.M.. Use the internediate Value Theorem to show there is a point on the path that the mond will cross at exactly the same time of day on both days. Hint: define u(t) to be the monk’s distance from the monastery, as a function of time, one the first day, and define d(t) to be his distance from the monatery, as a functino of time, on the second day. Let D be the distance from the monatery to the top of the mountain. Then use the given information to find u(0), u(12), d(0), d(12). And finally consider the function u − d and apply the intermediate value theorem on it. u(0) = 0, u(12) = D, d(0) = D, d(12) = 0. u and d are clearly continuous functions. Thus, so is u − d. Observe that (u−d)(0) = −D and (u−d)(12) = D. By the modified version of the intermediate value theorem there must be some time t0 between 0 and 12 such that (u − d)(t0 ) = 0 which is equivalent to u(t0 ) = d(t0 ). So at time t0 after 7:00 A.M. the mond will be at the same place on both days.
8. Find the slope of the tangent line to the graph of each function at the given point and determine an equation of the tangent line 2 (a) f (x) = 11x + 3 at ( 11 , 5)
(b) f (x) =
3 x
(c) f (x) =
2x (x+1)2
at (3, 1) at (0, 0)
2 (a) The equation is simply y = 11x + 3 at ( 11 , 5)
(b) f (x) = x3 at (3, 1) First the slope of the tangent at (3, 1) is m = lim
f (3+h)−f (3) = − 13 . h − 13 (x − 3) or y = − 31 x
h→0
Thus the equation of the tangeant line is y − 1 = (c) f (x) =
2x (x+1)2
at (0, 0)
First the slope of the tangent at (3, 1) is m = lim
h→0
f (0+h)−f (0) h
The equation of the tangeant line is y = 2x at (0, 0).
= 2.
+2
9. Let f = (x) = 3 − 2x + 4x2 and g(x) =
√
3x + 1
(a) Find the derivative f 0 (a) and g 0 (a), using the limit definition of the derivative. (b) Find the point on the graph of f where the tangent line to the curve is horizontal. Hint: A line is horizontal if its slope is null
g 0 (a)
= = = = = = = =
f 0 (a)
g(a + h) − g(a) h p √ 3(a + h) + 1 − 3a + 1 lim h→0 h p p √ √ ( 3(a + h) + 1 − 3a + 1)( 3(a + h) + 1 + 3a + 1) p lim √ h→0 h( 3(a + h) + 1 + 3a + 1) (3a + 3h + 1) − (3a + 1) lim p √ h→0 h( 3(a + h) + 1 + 3a + 1) 3h lim p √ h→0 h( 3(a + h) + 1 + 3a + 1) 3 lim p √ h→0 3(a + h) + 1 + 3a + 1 3 √ √ 3a + 1 + 3a + 1 3 √ 2 3a + 1 lim
h→0
= = = = = = =
f (a + h) − f (a) h 3 − 2(a + h) + 4(a + h)2 − (3 − 2a + 4a2 ) lim h→0 h 3 − 2a − 2h + 4a2 + 8ah + 4h2 − (3 − 2a + 4a2 ) lim h→0 h −2h + 8ah + 4h2 lim h→0 h h(−2 + 8a + 4h) lim h→0 h lim −2 + 8a + 4h lim
h→0
h→0
−2 + 8a
The derivative f 0 (a) = 2 + 8a is also the slope of the tangent line of f(x). For the second part of the question, we want to find a point (x0 , y0 ) such that the tangent line at this point is horizontal, i.e. f 0 (x0 ) = 0. Thus 2 + 8x0 = 0. So we find x0 = − 41 . Now yo = f (x0 ) = 3 + 21 + 41 = 14 . 4 At the point (− 41 , 14 ), the graph of f(x) has a horizontal tangeant line. 4 10. The number of bacteria after t hours in a controlled laboratory experiment is n = f (x). What is the meaning of the derivative f 0 (5) ? What are its units ? Suppose there is an unlimited amount of space and nutrients for the bacteria. Which do you think is larger f 0 (5) or f 0 (10)? If the supply of nutrients is limited, would that affect your conclusion ? Explain f 0 (5) is the rate of growth of the bacteria population when t = 5 hours. Its units are bacteria per hour. With unlimited space and nutrients, f’ should increase at t increases; so f 0 (5) < f 0 (10). If the supply of nutrients is limited, the growth rate slows down at some point in time, and the opposite may be true. 11. A particle moves along a straight line with equation of motion s = f (t) = 100 + 50t − 4.9t2 , where s is measured in meters and t in seconds. Find the velocity when t = 5 ?
f (t) = 100 + 50t − 4.9t2 . Using the limit definition of the derivative, compute f 0 (t) which is the velocity of the particle. f 0 (t) = lim −4.9h + 1 = 1m/s h→0
12. The total cost C(x) (in dollars) incurred by a company in manufacturing x surfboards a day is given by: C(x) = −10x2 + 300x + 130, 0 ≤ x ≤ 15. Find C 0 (x). What is the rate of change of the total cost when the level of production is ten surfboards a day ? C 0 (x) = −20x + 300 Rate of change of the total cost when the level of production is ten surfboards a day is simply C 0 (10) = −20 × 10 + 300 = 100.
