Congratulations to Andrei from Tudor Vianu National College, Bucharest,
Romania for this
excellent solution. (1) Plot the graph of the function y=f(x)
where f(x) = sinx +|sinx|. Differentiate the function and say where
the derivative is defined and where it is not defined.
I observe that sinx takes positive values between 2kp and
(2k+1)p (for integer k), that is in the first two quadrants,
and sinx takes negative values between (2k+1)p and
(2k+2)p, that is in the third and fourth quadrants. So
f(x)
= 2sinxfor 2kp £ x £ (2k+1)p
= 0 for (2k+1)p £ x £ (2k+2)p.
Below is
represented the graph of y = f(x):
Now I calculate the first derivative of f(x):
f¢(x)
= 2cosxfor 2kp < x < (2k+1)p
= 0 for (2k+1)p < x < (2k+2)p
The derivative
f¢(x) is not defined at the points x = kp for any integer k
and it does not have a tangent at these points.
(2) Now I express the function f(x) = sinx + cosx in the form
f(x)=Asin(x+a), find A and a and plot the graph of
this function. Similarly I express the function g(x) = sinx - cosx in the form g(x) = Bsin(x +b) where -p/2 < b < p/2, and plot its graph on the same axes.
f(x)
= sinx + cosx = Asin(x + a)
= Asin xcosa+ Acosx sina
As this formula must be valid
for any x, I obtain:
Asina = 1 and Acosa = 1
and hence tana = 1, a = p/4 and A=±Ö2.
Comment: If A has the meaning of an amplitude, A is positive,
and only the positive solution must be kept. This type of problem is
typical for the composition of oscillations.
Hence the graph of this function is a sine graph with a phase shift
of p/4, that is f(x)=0 when x=kp- p/4, it takes the
value 1 when x=2kp and x=2kp+ p/2 and the value -1 when
x=(2k+1)p and x=2kp- p/2, has maximum values (2kp +p/4, Ö2) and minimum values ((2k+1)p+ p/4, -Ö2)
In a similar manner I write g(x) = sinx - cosx:
f(x)
= sinx - cosx = Bsin(x + b)
= Bsinxcosb + Bcosx sinb
So I have
Bsinb = -1 and Bcosb = 1
and hence tanb = -1, b = -p/4 and B=Ö2. Hence
the graph of this function is a sine graph with a phase shift of
-p/4, that is f(x)=0 when x=kp+ p/4, it takes the value
1 when x=kp and x=2kp+ p/2, has maximum values (2k+1)p - p/4, Ö2) and minimum values (2kp- p/4, -Ö2).
(3)
Now, I calculate the function f(x) = sinx + |cosx| and plot the graph.
f(x)
= sinx + cosx = Ö2 sin(x + p/4) for 2kp-p/2 £ x £ 2kp+p/2
= sinx - cosx = Ö2 sin(x - p/4) for 2kp+ p/2 £ x £ 2kp+ 3p/2 .
The derivative is not defined at x = kp+ p/2 and
for other values of x it is:
