What type of graph does the equation #f(x)=sin(pi(x/(1+x^2)))# represent?

Question

Abigail Armstrong · Accepted Answer

Please see below.

Observe that #f(x)=sin(pi(x/(1+x^2)))# as #f(-x)=-x#, it is an odd function. Also #1+x^2# always exists and is positive and hence #x/(1+x^2)# depends on the sign of #x#. Further at #x=0#, #x/(1+x^2)=0# and as #f(x)=sin(pi(x/(1+x^2)))#, we have #f(0)=0#. Further at #x=+-1#, #x/(1+x^2)=+-1/2# and #f(+-1)=sin(pi/2)=+-1#. As maximum and minimum value of sine ratio is always #1# and #-1#, the function would never have a value in the interval #[-1,1]#. As the denoinator #1+x^2# rises a lot faster than #x# and hence as #x# increases from #1#, the value of #x/(1+x^2)# continues to decline and as #x# decreases from #-1#, the value of #x/(1+x^2)# continues to increase. As #x->+-oo#, we have #lim_(x->+-oo)sin(pi(x/(1+x^2)))=lim_(x->+-oo)sin(pi/(1/x+x))=sin0=0# and hence #f(x)=sin(pi(x/(1+x^2)))# continues to decline beyond value of #1# to #0# and also continues to increase beyond value of #-1# to #0#. To draw the graph we can select a few values of #x# such as #-5,-4,-3,-2,-1,0,1,2,3,4,5# and find #f(x)# for each to draw the graph. The graph appears as shown below. graph{sin(pi(x/(1+x^2))) [-5, 5, -2.5, 2.5]}

Oliver Atkinson · Answer

undefined The graph of the equation f(x) = sin(π(x/(1 + x^2))) represents a sinusoidal curve.