Math 170 Section 005

October 31, 2005 Worksheet

Solutions

f:=sqrt(3)*cos(x/2)+sin(x);

plot(f,x=-2*Pi..2*Pi);

(a)

f is increasing on (-5.2 +- 4kpi, -3.7 +- 4kpi), (-2.7 +- 4kpi, 1.2 +- 4kpi) for k=0,1,....

f is decreasing on (-3.7 +- 4kpi, -2.7 +- 4kpi), (1.2 +- 4kpi, 5.2 +- 4kpi) for k=0,1,....

(b)

df:=diff(f,x);

fsolve(df=0,x);

fsolve(df=0,x,x=-5..-3.5);

fsolve(df=0,x,x=-3..0);

fsolve(df=0,x,x=2..6);

f is increasing on (-5.182442973 +- 4kpi, -3.736974477 +- 4kpi), (-2.546210830 +- 4kpi, 1.100742334 +- 4kpi)

f is decreasing on (-3.736974477 +- 4kpi, -2.546210830 +- 4kpi), (1.100742334 +- 4kpi, 5.182442973 +- 4kpi)

(a)

sin(x) + C must be greater than zero.

(b)

If C > 1 then sin(x) + C will always be greater than zero and the graph is continuous everywhere.

(c)

If C <= 1 then sin(x) + C will always be less than or equal to zero and there will be no graph.

(d)

f:=C-> ln(sin(x)+C);

plot([f(-.5),f(0),f(.5),f(1),f(2),f(3),f(10)],x=-Pi..Pi,y=-10..5,legend=["C=-.5","C=0","C=.5","C=1","C=2","C=3","C=10"]);

(e)

y-intercept: (0,ln(C))

x-intercept: where sin(x) = 1-C

solve(f=0,x);

(f)

df:=diff(f(C),x);

Df:=unapply(df,C);

solve(Df(-.5)=0,x);

solve(Df(0)=0,x);

solve(Df(.5)=0,x);

solve(Df(1)=0,x);

solve(Df(2)=0,x);

solve(Df(3)=0,x);

solve(Df(10)=0,x);

f is decreasing on (pi/2 +- 2kpi, 3pi/2 +- 2kpi)

f is increasing on (-pi/2 +- 2kpi, pi/2 +- 2kpi)

(g)

There is a local minimum when x = -pi/2 +- 2kpi

There is a local maximum when x = pi/2 +- 2kpi

(h)

For each value of C f is increasing and decreasing on the same intervals and there are

local maximum and minimum at the same point, however the local maximum and

minimum values are different. In addition, the x and y intercepts change.

(i)

As C approaches infinity sin(x) becomes insignificant and the graph looks like ln(C).