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Minimum value as x varies

  • #1
hi, I am just tackling some of the harder questions we have to do and I came across this which I have never been told how to solve and my maths teacher too busy teaching the eaiser stuff to other people to expalin this lol, annoying i know, a few of us are in this situation, anyways, hear it is, thnx.

Homework Statement



The expression x^2 - 2mx has a minimum value as x varies

qu) Find the minimum value of x^2 - 2mx
Give your answer in terms of m

Homework Equations





The Attempt at a Solution



I have no idea, how to approach this. I have tried reasoning logically how it might work, but came nowhere as I dunno how the expression can have a minimum value :S is the power of 2 a clue? A few of us (in my class) have tried but no-one has the fainted idea. hope you can help

thnx
 

Answers and Replies

  • #2
236
0
If a function have a maximum or minimum at a certain point, what can you say about the derivative at that point?




Note that this is the PRECALCULUS fourm. Knowledge of derivatives should not be expected
Integral
 
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  • #3
what's a derivative?
 
  • #4
236
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A derivative is the rate of change of a quantity. A derivative is an instantaneous rate of change: it is calculated at a specific instant rather than as an average over time.
 
  • #5
Dick
Science Advisor
Homework Helper
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618
There is a more elementary way to do problems like this. Complete the square in the expression. Can you handle this? Ie write it as (x-m)^2+SOMETHING. Can you figure out what the SOMETHING is?
 
  • #6
312
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There is a more elementary way to do problems like this. Complete the square in the expression. Can you handle this? Ie write it as (x-m)^2+SOMETHING. Can you figure out what the SOMETHING is?
You can do this or find the axis of symmetry and vertex. For parabolas, you dont need to use derivatives
 
  • #7
1,341
3
You can do this or find the axis of symmetry and vertex. For parabolas, you dont need to use derivatives
By completing the square you're essentially doing that, it will be in vertex form. [tex]y=a(x-h)^2+k[/tex]

Whatever makes (x-h) equal 0 is your x-coordinate, k is your y-coordinate for a min/max of a quadratic

Or at least I believe what I said to be correct, maybe I'm wrong, dunno :-/
 
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  • #8
286
0
if you change the value of "m", with it always be a "happy parabola"?

How do you find the axis of symmetry for the following parabolas?
y=x^2 + 4x -8
y=x^2 + 4x

Then, how would you find the axis of symmetry for your parabola?

Notice that the lowest point for a "happy" parabola is always on this axis. (likewise for a "sad" parabola)
 
  • #9
the lowest point is min value let say ax^2+bx+c
the lowest point x-coordinate is /frac{b}{-2a}
 
  • #10
Gib Z
Homework Helper
3,346
5
Nice try at the latex dragonlorder, welcome to Pf. You forgot [ tex ] brackets though. As said before, the x value at the lowest point is [itex]\frac{-b}{2a}[/itex], to get the lowest value, they are asking for the Y value! So dont forget to substitute your x value back into your equation and get the y value.
 
  • #11
There is a more elementary way to do problems like this. Complete the square in the expression. Can you handle this? Ie write it as (x-m)^2+SOMETHING. Can you figure out what the SOMETHING is?
would it be (x-m)^2 + m^2?

where would i go from there?
 
  • #12
Dick
Science Advisor
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I would say x^2-2mx=(x-m)^2-m^2. Now (x-m)^2 is always greater than or equal to zero, yes? So what is the minimum value?
 
  • #13
o yeh oops, lol, my bad

would the minimum value be -m^2 then?
 
  • #14
Dick
Science Advisor
Homework Helper
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You are so right.
 
  • #15
thnx for all the help buddy :D
 

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