Question

Quadratics and functions, Help please ?

Answer 1

`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.