Find the maximum or minimum values of the function.
`y=9-(x-3)^(2)`

Explanation

`y = 9 - (x - 3)^(2) = 9 - x^(2) - 9 + 6x`
or `y = 6x - x^2`
`:. (dy)/(dx) = 6 - 2x`
minimum or maximum value of y we will substitue `(dy)/(dx) = 0`
or, `6 - 2x = 0" "x = 3`
To check whether value of y is maximum or minimum at x = 3 we will have to check wheter `(d^2y)/(dx^2)` is possitive or negative.
`(d^2 y)/(dx^2) = -2`
or `(d^2y)/(dx^2)` is negative at x = 3. Hence, value of y is maximum. This maximum value of y is,
`y_("max") = 9 - (3 -3)^(2) = 9`