Minimum value as x varies

  • #1
Trail_Builder
149
0
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
ziad1985
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
 
Last edited by a moderator:
  • #3
Trail_Builder
149
0
what's a derivative?
 
  • #4
ziad1985
236
0
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
26,263
619
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
turdferguson
312
0
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
Feldoh
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 :-/
 
Last edited:
  • #8
drpizza
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
dragonlorder
18
0
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
6
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
Trail_Builder
149
0
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
Homework Helper
26,263
619
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
Trail_Builder
149
0
o yeh oops, lol, my bad

would the minimum value be -m^2 then?
 
  • #14
Dick
Science Advisor
Homework Helper
26,263
619
You are so right.
 
  • #15
Trail_Builder
149
0
thnx for all the help buddy :D
 

Suggested for: Minimum value as x varies

Replies
21
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
5
Views
588
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
16
Views
3K
Replies
1
Views
1K
Top