2) You should draw F(t) increasing and concave up for a while, then
increasing and concave down, leveling off after a few minutes. It should
also be the case that F(0) is about 55 or so, and it should probably level
off around 110. Possible graphs (with t in seconds) are shown below.
![]()
![[Graphics:ex2ansgr3.gif]](http://www.central.edu/homepages/lintont/classes/spring00/Calculus/exam2/ex2ansgr3.gif)
![]()
3) f '(3) = ![]()
= ![]()
= ![]()
=
13
+ 2h = 13 + 0 = 13.
4) nderiv(2.3^X, X, -1) = 0.362134
5) given f '(x) graph.
a) Concave down as f '(x) is decreasing at x = 0.
b) For x = 0 to 1.9, f '(x) is negative, so f(x) is decreasing.
c) At x = 0, since f '(x) is positive for the entire time, f(x) just
keeps getting bigger.
d) x = -2 is a chair seat. x = 0 is a hilltop and x = 1.9 is a valley
bottom.
e) At x = -2, -0.75 and 1.25, the hills and valleys of f '(x) are inf
pts of f(x).
f) The tangent line to f '(x) at x = 0 has a slope of about -6.4
g) y = -8(x - 1) + 4
6) f(x) =
a) This is just
=
1.5.
b) Using nderiv, I get roughly f '(0) = 0.75.
c) The limits are:
i) -2/3
ii) -20 / 3
iii) -infinity
7) Calculus books and heights.
a) v(t) = 96 - 32t
b) 48 ft / sec
c) At t = 3 the book is 144 feet in the air.
d) s(t) = 6 (for the second time) at t = 5.93684 seconds.
8) Given a graph of f(x).
a) Negative as f(x) is decreasing at x = 1.
b) concave down.
c) Decreasing as f(x) is concave down there.
d) The limit is the defintion of f '(1), which, via a tangent line
slope is about equal to -1.33.
e) Draw the line through (0.5, f(0.5) ) and (1.5, f(1.5) ).
f) Via a tangent line slope, I get f '(0.5) = 0.75.
g) Positive, since f(x) is concave up at x = 1.9.
h) The plot below is what you should get.
![]()