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 March 18th, 2017, 10:45 PM #1 Senior Member   Joined: Jan 2017 From: US Posts: 104 Thanks: 5 How to Find a Minimum/Maximum in a Function? What are the minimums and maximums in a function, and what are their purpose? How would you pick them out of a function? Here is a function if you want to use it as an example: f(x)=-3x$\displaystyle ^2$+1-x
 March 18th, 2017, 10:50 PM #2 Member   Joined: Mar 2017 From: Tasmania Posts: 36 Thanks: 2 You can complete the square to find the min and max but I believe there is also a formula you can use Thanks from Indigo28
March 18th, 2017, 11:24 PM   #3
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 Originally Posted by Indigo28 What are the minimums and maximums in a function, and what are their purpose? How would you pick them out of a function? Here is a function if you want to use it as an example: f(x)=-3x$\displaystyle ^2$+1-x
A parabola either has a minimum, if it is facing upwards, or a maximum, if it is facing downwards. This point will always be the vertex of parabola.

You can tell whether it is facing upwards or downwards by looking at the sign of the coefficient of the x^2 term.

A positive coefficient means the parabola faces upwards.

Negative means it faces downwards.

So if we compete the square and put the function into vertex form we can

a) pick the vertex out, and you know this will be either the min or max

b) examine the coefficient of $(x-h)^2$ to detemine it's sign and whether the point in (a) is a min or a max.

$f(x) = -3x^2 - x +1 =$

$-3\left(x^2 + \dfrac x 3 - \dfrac 1 3\right)=$

$-3\left(\left(x+\dfrac 1 6\right) - \left(\dfrac 1 6 \right)^2 - \dfrac 1 3\right) =$

$-3\left(\left(x+\dfrac 1 6\right)^2 -\dfrac {13}{36}\right) =$

$-3\left(x+\dfrac 1 6\right)^2 + \dfrac{13}{12}$

We see that the $(x-h)^2$ term has coefficient $-3$ and thus the vertex, $\left(-\dfrac 1 6,\dfrac {13}{12}\right)$ is a maximum.

March 19th, 2017, 01:17 AM   #4
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 Originally Posted by romsek A parabola either has a minimum, if it is facing upwards, or a maximum, if it is facing downwards. This point will always be the vertex of parabola. You can tell whether it is facing upwards or downwards by looking at the sign of the coefficient of the x^2 term. A positive coefficient means the parabola faces upwards. Negative means it faces downwards. So if we compete the square and put the function into vertex form we can a) pick the vertex out, and you know this will be either the min or max b) examine the coefficient of $(x-h)^2$ to detemine it's sign and whether the point in (a) is a min or a max. $f(x) = -3x^2 - x +1 =$ $-3\left(x^2 + \dfrac x 3 - \dfrac 1 3\right)=$ $-3\left(\left(x+\dfrac 1 6\right) - \left(\dfrac 1 6 \right)^2 - \dfrac 1 3\right) =$ $-3\left(\left(x+\dfrac 1 6\right)^2 -\dfrac {13}{36}\right) =$ $-3\left(x+\dfrac 1 6\right)^2 + \dfrac{13}{12}$ We see that the $(x-h)^2$ term has coefficient $-3$ and thus the vertex, $\left(-\dfrac 1 6,\dfrac {13}{12}\right)$ is a maximum.
THANK YOU!! This was super helpful and exactly what I needed. Seriously, thanks for taking the time to explain all of this.

March 31st, 2017, 02:40 AM   #5
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 Originally Posted by Indigo28 THANK YOU!! This was super helpful and exactly what I needed. Seriously, thanks for taking the time to explain all of this.

You could also graph this equation if you have something like a Ti-89 and hit F5 and select either minimum (if pointing up) or maximum (if pointing down) then go to the left side (lower bound) press enter and right side of the curvature (upper bound) (or go right then left, doesn't matter), hit Enter and boom there it is.

You could find the vertex a number of ways. Firstly, find x=-b/2a and plug that x value in to get y afterwards and that point is your vertex. Or you could put the equation in vertex form by completing the square.

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