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March 18th, 2017, 09:45 PM   #1
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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
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March 18th, 2017, 09:50 PM   #2
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You can complete the square to find the min and max but I believe there is also a formula you can use
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March 18th, 2017, 10:24 PM   #3
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Quote:
Originally Posted by Indigo28 View Post
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.
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March 19th, 2017, 12:17 AM   #4
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Quote:
Originally Posted by romsek View Post
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.
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