Solutions to Exam I, Math 151, Spring 1996
Part I - Multiple Choice
1 c 2 b (should be +2) 3 d 4 c 5 d 6 e 7 e 8 b
9 b 10 e 11 a 12 a
Part II - Work Out
13. a)
> f:=cos(3*x+x^2); der_f:=diff(f,x);
2
f := cos(3 x + x )
2
der_f := - sin(3 x + x ) (3 + 2 x)
b)
> g:=(x^4-3*x)^4; der_g:= diff(g,x);
4 4
g := (x - 3 x)
4 3 3
der_g := 4 (x - 3 x) (4 x - 3)
14. a) Note: This material has not been covered yet for fall of 1996
> h:=(3*x)/sin(x); der_h:=diff(h,x);
x
h := 3 ------
sin(x)
3 x cos(x)
der_h := ------ - 3 --------
sin(x) 2
sin(x)
b)
> k:=t^2*(t^3-t)^(3/2); der_k:=diff(k,t);
2 3 3/2
k := t (t - t)
3 3/2 2 3 1/2 2
der_k := 2 t (t - t) + 3/2 t (t - t) (3 t - 1)
15. a) avg vel t=1 to t=3 is 11.3 meters/sec.
b) inst. vel at t=3 is -2.7 meters/sec
c) max height is 46.1 meters
16. A tangent line to the y=x^2 that contains (0,-4) is y=4x-4,
(this line is tangent at a=2). Another tangent line is y=-4x-4
(which is tangent at a=-2).
17. a) Domain of the derivative of f is all real numbers x
except x=-2 and x=4.
b) The slopes of the line segments in the graph of f are
3, -2 and 6. Thus the graph of the derivative of
f is a horizontal line at height 3 for x <-2; then
a horizontal line at height -2 for -2 < x < 4; followed
by a horizontal line at height 6 for 4< x.
> p1:=plot(3,x=-4..-2): p2:=plot(-2,x=-2..4): p3:=plot(6,x=4..7):
> with(plots): display({p1,p2,p3});
c) Range of the derivative of f is {-2, 3, 6}
18. The limit of (x-2)*sin(1/(x-2)) as x -> 2 is the same
as the limit of t*sin(1/t) as t->0, which is zero by
the squeeze theorem (|t*sin(1/t)| <= |t| which ->0
as t->0).
19. The expression given factors as
>
> abs(x-3)*abs(x+2)/((x-3)*(x+1));
|(x - 3)| |(x + 2)|
---------------------
(x - 3) (x + 1)
As x-> 3 from the left, x-3 is the negative of |x-3|.
Therefore the factor |x-3|/(x-3) equals -1.
The other factor |x+2|/(x+1) -> 5/4 as x ->3.
So the answer is -5/4.