cal

on what intervals is f increasing justify your answer

(-6,-2) u (2.5) bc f'(x) > 0 on the intervals

for t > 0 the postilion of a particle moving along the x-axis is given by x(t) = sin t - cos t. what is the acceleration of the particle at the point where the velocity is first equal to 0

(A) -root2

the function f is continuous on the closed interval [0,2] and has values that are in the table above. The equation f(x) 1/2 must have at least two solutions in the interval [0,2] if k=

(A) 0

what is lim h->0 cos(3pi/2 +h) - cos(3pi/2)/h ?

(A) 1

what is lim x approaches infinity x^2-4/2+x-4x^2

(B) -1/4

if f(x)= sin^2 (3-x) then f'(0) =

(B) -2sin 3 cos3

lim h->0 ln(4+h)-ln(4)/h is

(B) 1/4

limx -> pi: cosx + sin(2x) +1/x^2-pi^2 is

(B) 1/pi

lim x-> 0 : 7x-sinx/x^2+sin(3x)

(B) 2

lim as x approaches infinity sqaure root 9x^4 +1/x^2-3x+5

(B) 3

at which of the 5 points on the graph in the figure at the right are dy/dx and d2y/dx2 both negative

(B) B

let f be the function given by f(x)=300x-x^3 on which of the following intervals is the functions f increasing

(B) [-10,10]

if f(x) = sin(ln(2x)) then f'(x) =

(B) cos(ln(2x))/x

if f is continuous for a< x < b and differentiable for a< x < b which of the following could be false

(B) f'(c)=0 for some c such that a<c<b

for time t > 0 th e velocity of a particle moving along the x-axis is given by v(t)=(t-5)(t-2)^2. At what values of t is the acelartion of the particle equal to 0

(C) 2 and 4

the graph of the function f is shown in the figure above. for how many values of x in the open interval (-4,4) is f discontinuous

(C) 3

Three graphs labeled I,II,and III are shown above. One is the graph of f f' and f'' which of the following correctly identifies each of the three graphs

(C) II III I

The graph of the function j is shown above. Which of the following statements is false?

(C) lim f(x) x approaches 4 exists.

the graph of the piecewise-defined function f is shown in the figure above. The graph has a vertical tangent at x=-2 adn horizontal tangent lines at x=-3 and x=-1. What are all values of x,-4 x 3 at which f is cont but not differentiable?

(C) x=-2 and x= 1

the cost in dollars o shred the confidential documents is moddled by C, a differ function of the weight of documents in pounds. Of the following, which is the best interpretation of C'(500)=80

(D) the cost to shred documents is increasing at a rate of $80 per pound when the weight of the documents is 500 pounds

the possition of a particle moving along a line is given by s(t)= 2t^3 - 24t^2 +90t +7 for t>0. for what values of t is the speed of the particle increasing

(E) 3<t<4 and t>5

if y=(x^3 - cosx)^5 then y' =

(E) 5(x^3 -cosx)^4 (3x^2 +sinx)

If f(x)= 7x-3+ln x then f'(1)=

(E) 8

if y= xsinx then dy/dx =

(E) x(sinx-cosx)

the graph of a function f is hown above if lim as x appraoches b f(x) exsists and f is not cont at b, then b=

(b) 0

the graph of the functions f and g are shown above. the value of lim as x approaches 1 f(g(x))is

(c) 3

is there a time t 2<t<4 at which c'(t) =2 justify your answer

12.8-8.8/4-2=2 c is differ so c must be cont on 2<t<4 therefore mvt can be applied so there is at least 1 time at which c'(t) =2

for each of the f''(-5) and F''(3) find the values or explain why it does not exsist

f''(-5) = 1-0/-6--2= -1/2 and the lim of f''(3) are not equal therefore DNE

th efinction g is defined by g(x)= (f(x))^3 if f(3)=-5/2 find the slope of the line tangent to the graph of g at x=3

g'(3) = -75

explain why there must be a value c for 1<c<3 such that h'(c)= -5

since H(x) is cts on the closed interval 1,3 and h(x) is diff on (1.3) h'(c) =h(3) - h(1)/3-1 -5=-7-3/2

explain why there must be a value r for 1<r<3 such that h(r)=-5

since h(x) is cts on a closed interval 1,3 by IVT h(1)=3 h(3)=-7 h(r) =-5 since -5 is between h(1) and h(3)

on what open interval contained in 0<x<8 is the graph of f both con down and inc explain

at x=0,1 and 3,4 f is both cd and inc bc f' is pos and dec

find all the values of x on the open interval 0<x<8 for which the finction f has a local minimum. justify your answer

at x=6 there is a min bc f' chnages from neg to pos

use the data in the table to approximate c'(3.5) show the computations that lead to your answer, and indicate units of measure

c(4)-c(5)/4-3= 1.6 ounces / min