# Function.What are the extreme values of the function f(x) = 2x-3x^2 + 5.

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The give function f(x) = 2x - 3x^2 + 5. At the extreme point f'(x) = 0

f'(x) = 2 - 6x

2 - 6x = 0

=> x = 1/3

f(1/3) = 2*(1/3) - 3*(1/3)^2 + 5

=> 2/3 - 1/3 + 5

=> 1/3 + 5

=> 16/3

**The extreme value of the function is 16/3**

We'll re-arrange the terms of the equation of the function:

f(x) = -3x^2 + 2x + 5

We notice that the coefficient of x^2 is a = -3, so the vertex of the parabola is a maximum point.

We'll calculate the coordinates of the vertex using the identities:

xV = -b/2a

yV = -delta/4a

a = -3, b = 2 , c = 5

xV = -2/-6

xV = 1/3

yV = -(b^2 - 4ac)/4a

yV = (-60 - 4)/-12

yV = -64/-12

yV = 16/3

**The extreme point of the given function is a maximum point represented by the following coordinates (1/3 ; 16/3).**