In 18, each function is continuous and defined on a closed interval. It therefore

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = 2x 1, 0 x 1

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = x2 + 1, 1 x 1

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = sin(2x), 0 x

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = cos x 2 , 0 x 2

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = |x|, 1 x 1

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = (x 1)2(x + 2), 2 x 2

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = e|x|, 1 x 1

In Problems 18, each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) f (x) = ln(x + 1), 0 x 2

Sketch the graph of a function that is continuous on the closed interval [0, 3] and has a global maximum at the left endpoint and a global minimum at the right endpoint.

Sketch the graph of a function that is continuous on the closed interval [2, 1] and has a global maximum and a global minimum in the interior of the domain of the function.

Sketch the graph of a function that is continuous on the open interval (0, 2) and has neither a global maximum nor a global minimum in its domain.

Sketch the graph of a function that is continuous on the closed interval [1, 4], except at x = 2, and has neither a global maximum nor a global minimum in its domain.

In Problems 1318, use a graphing calculator to determine all local and global extrema of the functions on their respective domains. f (x) = 3 x, x [1, 3)

In Problems 1318, use a graphing calculator to determine all local and global extrema of the functions on their respective domains. f (x) = 5 + 2x, x (2, 1)

In Problems 1318, use a graphing calculator to determine all local and global extrema of the functions on their respective domains. f (x) = x2 2, x [1, 1]

In Problems 1318, use a graphing calculator to determine all local and global extrema of the functions on their respective domains. f (x) = (x 2)2, x [0, 3]

In Problems 1318, use a graphing calculator to determine all local and global extrema of the functions on their respective domains. f (x) = x2 + 1, x [2, 1]

In Problems 1318, use a graphing calculator to determine all local and global extrema of the functions on their respective domains. f (x) = x2 x, x [0, 1] c f _ (c) = 0 f (x) x = c

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = x2

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = (x 4)2

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = x2

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = (x + 3)2

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = x3

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = x5

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = (x + 1)3

In Problems 1926, find c such that f  (c) = 0 and determine whether f (x) has a local extremum at x = c. f (x) = (x 3)5

Show that f (x) = |x| has a local minimum at x = 0 but f (x) is not differentiable at x = 0.

Show that f (x) = |x 1| has a local minimum at x = 1 but f (x) is not differentiable at x = 1.

Show that f (x) = |x2 1| has local minima at x = 1 and x = 1 but f (x) is not differentiable at x = 1 or x = 1.

Show that f (x) = |x2 4| has local maxima at x = 2 and x = 2 but f (x) is not differentiable at x = 2 or x = 2.

Graph f (x) = __ 1 |x| __ , 1 x 2 and determine all local and global extrema on [1, 2].

Graph f (x) = __ |x| 2 __ , 3 x 3 and determine all local and global extrema on [3, 3].

Suppose the size of a population at time t is N(t) and its growth rate is given by the logistic growth function dN dt = rN _ 1 N K _ , t 0 where r and K are positive constants. (a) Graph the growth rate dN dt as a function of N for r = 2 and K = 100, and find the population size for which the growth rate is maximal. (b) Show that f (N) = rN(1 N/K), N 0, is differentiable for N > 0, and compute f _ (N). (c) Show that f _ (N) = 0 for the value of N that you determined in (a) when r = 2 and K = 100.

Suppose that the size of a population at time t is N(t) and its growth rate is given by the logistic growth function dN dt = rN _ 1 N K _ , t 0 where r and K are positive constants. The per capita growth rate is defined by g(N) = 1 N dN dt (a) Show that g(N) = r _ 1 N K _ (b) Graph g(N) as a function of N for N 0 when r = 2 and K = 100, and find the population size for which the per capita growth rate is maximal.

Suppose f (x) = x2, x [0, 2]. (a) Find the slope of the secant line connecting the points (0, 0) and (2, 4). (b) Find a number c (0, 2) such that f _ (c) is equal to the slope of the secant line you computed in (a), and explain why such a number must exist in (0, 2).

Suppose f (x) = 1/x, x [1, 2]. (a) Find the slope of the secant line connecting the points (1, 1) and (2, 1/2). (b) Find a number c (1, 2) such that f _ (c) is equal to the slope of the secant line you computed in (a), and explain why such a number must exist in (1, 2).

Suppose that f (x) = x2, x [1, 1]. (a) Find the slope of the secant line connecting the points (1, 1) and (1, 1). (b) Find a number c (1, 1) such that f _ (c) is equal to the slope of the secant line you computed in (a), and explain why such a number must exist in (1, 1).

Suppose that f (x) = x2 x 2, x [1, 2]. (a) Find the slope of the secant line connecting the points (1, 0) and (2, 0). (b) Find a number c (1, 2) such that f _ (c) is equal to the slope of the secant line you computed in (a), and explain why such a number must exist in (1, 2).

Let f (x) = x(1 x). Use the MVT to find an interval that contains a number c such that f _ (c) = 0.

Let f (x) = 1/(1 + x2). Use the MVT to find an interval that contains a number c such that f _ (c) = 0.

Suppose that f (x) = x2 + 2. Explain why there exists a point c in the interval (1, 2) such that f _ (c) = 1.

Suppose that f (x) = x3. Explain why there exists a point c in the interval (1, 1) such that f _ (c) = 1.

Sketch the graph of a function f (x) that is continuous on the closed interval [0, 1] and differentiable on the open interval (0, 1) such that there exists exactly one point (c, f (c)) on the graph at which the slope of the tangent line is equal to the slope of the secant line connecting the points (0, f (0)) and (1, f (1)).Why can you be sure that there is such a point?

Sketch the graph of a function f (x) that is continuous on the closed interval [0, 1] and differentiable on the open interval (0, 1) such that there exist exactly two points (c1, f (c1)) and (c2, f (c2)) on the graph at which the slope of the tangent lines is equal to the slope of the secant line connecting the points (0, f (0)) and (1, f (1)). Why can you be sure that there is at least one such point?

Suppose that f (x) = x2, x [a, b]. (a) Compute the slope of the secant line through the points (a, f (a)) and (b, f (b)). (b) Find the point c (a, b) such that the slope of the tangent line to the graph of f at (c, f (c)) is equal to the slope of the secant line determined in (a). How do you know that such a point exists? Show that c is the midpoint of the interval (a, b); that is, show that c = (a + b)/2.

Assume that f is continuous on [a, b] and differentiable on (a, b). Show that if f (a) < f (b), then f _ is positive at some point between a and b.

Assume that f is continuous on [a, b] and differentiable on (a, b). Assume further that f (a) = f (b) = 0 but f is not constant on [a, b]. Explain why there must be a point c1 (a, b) with f _ (c1) > 0 and a point c2 (a, b) with f _ (c2) < 0.

A car moves in a straight line. At time t (measured in seconds), its position (measured in meters) is s(t) = 1 10 t2, 0 t 10 (a) Find its average velocity between t = 0 and t = 10. (b) Find its instantaneous velocity for t (0, 10). (c) At what time is the instantaneous velocity of the car equal to its average velocity?

A car moves in a straight line. At time t (measured in seconds), its position (measured in meters) is s(t) = 1 100 t3, 0 t 5 (a) Find its average velocity between t = 0 and t = 5. (b) Find its instantaneous velocity for t (0, 5). (c) At what time is the instantaneous velocity of the car equal to its average velocity?

Denote the population size at time t by N(t), and assume that N(0) = 50 and |dN/dt| 2 for all t [0, 5]. What can you say about N(5)?

Denote the biomass at time t by B(t), and assume that B(0) = 3 and |dB/dt| 1 for all t [0, 3].What can you say about B(3)?

Suppose that f is differentiable for all x R and, furthermore, that f satisfies f (0) = 0 and 1 f _ (x) 2 for all x > 0. (a) Use Corollary 1 of the MVT to show that x f (x) 2x for all x 0. (b) Use your result in (a) to explain why f (1) cannot be equal to 3. (c) Find an upper and a lower bound for the value of f (1).

Suppose that f is differentiable for all x R with f (2) = 3 and f _ (x) = 0 for all x R. Find f (x).

Suppose that f (x) = e|x|, x [2, 2]. (a) Show that f (2) = f (2). (b) Compute f _ (x), where defined. (c) Show that there is no number c (2, 2) such that f _ (c) = 0. (d) Explain why your results in (a) and (c) do not contradict Rolles theorem. (e) Use a graphing calculator to sketch the graph of f (x).

Use Corollary 2 of the MVT to show that if f (x) is differentiable for all x R and satisfies | f (x) f (y)| |x y|2 (5.3) for all x, y R, then f (x) is constant. [: Show that (5.3) implies that lim xy f (x) f (y) x y = 0 (5.4) and use the definition of the derivative to interpret the left-hand side of (5.4).]

We have seen that f (x) = f0erx satisfies the differential equation d f dx = r f (x) with f (0) = f0. This exercise will show that f (x) is in fact the only solution. Suppose that r is a constant and f is a differentiable function, d f dx = r f (x) (5.5) for all x R, and f (0) = f0. The following steps will show that f (x) = f0erx , x R, is the only solution of (5.5). (a) Define the function F(x) = f (x)erx , x R Use the product rule to show that F_ (x) = erx [ f _ (x) r f (x)] (b) Use (a) and (5.5) to show that F_ (x) = 0 for all x R. (c) Use Corollary 2 to show that F(x) is a constant and, hence, F(x) = F(0) = f0. (d) Show that (c) implies that f0 = f (x)erx and therefore, f (x) = f0erx