What are the extreme values of the function f(x) = 2x-3x^2 + 5.
Given the parabola f(x) = 2x- 3x^2 + 5.
We need to determine the extreme values of the function.
We know that the extreme values are the values of f(x) such that the function has maximum or minimum points.
To find extreme values, we need to differentiate f(x) and determine the critical values.
==> f(x) = -3x^2 + 2x + 5
==> f'(x) = -6x + 2
The critical values are the derivative's zeros.
==> -6x + 2 = 0
==> -6x = -2
==> x = -2/-6 = 1/3
Then the function has an extreme value when x= 1/3
==> f(1/3) = -3 ( 1/3)^2 + 2( 1/3) + 5
= -1/3 + 2/3 + 5
= ( -1 + 2 + 15) /3 = 16/3
Since the sign of x^2 is negative, the the function has a maximum point.
Then, the maximum point is f(1/3) = 16/3.
We notice that the equation is a quadratic and the critical point of a quadratic is represented by the vertex of the parabola.
We'll re-write 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 is a maximum point.
We'll calculate the coordinates of the vertex using the formula:
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 and it has the coordinates (1/3 ; 16/3).