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`f = 6x -x^2`
`f' = 6-2*x`
`f' = 2(3-x)`
At extrema, f'=0, this give `3- x = 0 therefore x =3 `
At x = 3, there is a local extreme point. At x = 3,
`f = 6*3-3^2 = 18-9 = 9`
Second derivative test,
`f'' = 2(0-1) = -2`
`f'' lt 0` for all `x` , therefore at `x =3, f'' lt 0.`
This suggests, the extrem point at x = 3 is a maximum.
Therefore, there is a local maximum point of (3,9)
Find the first derivative.
Now we need to find the value(s) of x that make the first derivative zero to find the critical points.
`x=3 ` (critical point)
This will be a relative extrema if it changes sign, so find 2 values around it to test using the first derivative test, like 2 and 4
Since the sign changes then the point `(3,9) ` is a relative extrema
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