Determine the maximum of the function x^2-x-6=0.
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The vertex is the maximum/minimum point of a parabola.
To find the vertex:
x^2-x-6=0
==> (x-1/2)^2-1/4-6=0
Notice if we eliminate the parentheses, the equation becomes:
x^2-x+1/4-1/4-6=0. Therefore, the value of the equation has not changed.
(x-1/2)^2-6.25=0.
The vertex of the equation is 1/2, -6.25
However, since the parabola is concave up, the vertex gives only the minimum value. To find the maximum point of the function, we need certain range of x value. If there is no range, then the y-value of function goes to infinity, thus has no maximum point.
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To determine the maximum of f(x) = x^2-x-6.
f(x) = x^2-x+(1/2)^2 -(1/2)^2-6.
=> f(x) = (x-1/2)^2 -6.25.
=> f(x) > = -6.25.
Therefore f(x) >= -6.25 for all x as (x-1/2) ^2 > = 0.
So minimum of of f(x) = -6.25 when x= 1/2.
Also f(x) becomes infinite as x--> infinity or x--> -infinity. So there is no maximum for f(x).
To determine the maximum of the given function means to find the vertex of a parabola.
We'll write the function as:
f(x) = a(x-h)^2 + k, where the vertex has the coordinates v(h,k)
We'll re-write the given function:
f(x) = 1(x^2 - 1x) - 6
We'll complete the square:
x^2 -2*(1/2) x + (1/2)^2 = (x - 1/2)^2
So, we'll add and subtract the value 1/4:
f(x) = 1(x^2 - x + 1/4) - 1/4 - 6
We'll combine like terms outside the brackets:
f(x) = (x - 1/2)^2 - 25/4
We'll compare the result with the standard form and we'll get the coordinates of the vertex:
(x - 1/2)^2 - 25/4 = a(x-h)^2 + k
h = 1/2
k = -25/4
The coordinates of the vertex are:V (1/2 ; -25/4)
Another way to determine the local extreme of the function is to use the first derivative of the function (the vertex is a local extreme).
f'(x) = 2x - 1
We'll determine the critical value of x:
2x - 1 = 0
2x = 1
x = 1/2
Now, we'll calculate the y coordinate of the local extreme:
f(1/2) = (1/2)^2 - 1/2 - 6
f(1/2) = 1/4 - 1/2 - 6
f(1/2) = (1-2-24)/4
f(1/2) = -25/4
We notice that the vertex of the function is located in the 4th quadrant.
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