Quadratics and functions, Help please ?

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1 Answer
May 5, 2018

f(0) = 1
Range = [-5,1]
k = 0.84
g^(-1)(x) = 1/(3cos(x) - 2)

Explanation:

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.