Quadratics and functions, Help please ?

Question

Daniel Acosta · Accepted Answer

#f(0) = 1#
Range #= [-5,1]#
#k = 0.84#
#g^(-1)(x) = 1/(3cos(x) - 2)# The solution at a point is found by substituting the value for the variable and calculating the result. #3cos(x) - 2# ; #3cos(0) - 2 = 3 - 2 = 1# The "range" of a function is the spread from its lowest to highest value. In this case the domain is given as #0 <= x <= 2pi# so the range is the minimum and maximum values of the function at those points. It will be a minimum when #cos(x) = -1# and maximum when #cos(x) = 1#. Range #= [-5,1]# Sketch the curve by making a short table of #x and f(x)#, then plot them on graph paper or use a computer program. It will be an off-set and expanded cosine curve. For a function to have an inverse, the original must not be equal to zero. So, the highest value in a range meeting that criterion is the answer. The expression of 'g' is: #g^(-1)(x) = 1/(3cos(x) - 2)# It is undefined at #3cos(x) - 2 = 0#, so #cos(x) = 2/3# is the 'limit'. #cos(0.84) = 2/3# in radians. Because the function is cyclic, I am not sure what is meant by "highest" value of k. I could be anything except where #cos(x) = 2/3#. So, yes, if we look at the continuous function, the maximum k would be a multiple of #pi#. I took it as the highest point reached before the function was not invertable.