need help in maths!
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19 May 2010, 17:50
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Journals
hello guys, i need ur maths help, since im nob there and dont understand actuell stuff.
We have the function f(x)=5x²+x-16
I have to find the minima. Plz help!
Thanks in advance
ps: http://www.youtube.com/watch?v=dnLjr8TaE1Q
We have the function f(x)=5x²+x-16
I have to find the minima. Plz help!
Thanks in advance
ps: http://www.youtube.com/watch?v=dnLjr8TaE1Q
= win?
=
6x³-16 + 6x²f-16
=
6x³ + 6x²f -32
thanks for getting me back 4 years ago : D
->0=10x+1
x=-0.1
in f(x)
y=5x(-0.1)²+(-0.1)-16
y=-16.05
->P(-0.1|-16.05)
also, wenne die erste ableitung gleich 0 setzt
deswegen setzt du die 1. ableitung gleich 0..um die punkte rauszufinden, in denen der graph keine steigung hat...
eine tangente ist eine gerade die einen bestimmten punkt des graphen berührt aka tangiert, im matheunterricht meist den wendepunkt
wenn du aber z.b. ne tangente an den extrempunkt legen würdest hätte diese die steigung m=f'(x)=0
an 3-4tagen ne stunde sollte da locker reichen
http://gratis-tube.de/data/200807/tangente.png
die rote linie ist die tangente...
wenn du eine solche tangente an einen minimal/maximal punkt legst, ist diese tangente waagerecht...und hat dementsprechend keine steigung...
http://www.mathe-schumann.de/veroeffentlichungen/extrem/Image24.gif
jetzt muss ich mir das alles irgendwie merken...
SO viel Info auf einmal. :(
dann auflösen nach x.
+wegen dem positiven x² kanns ja nur minimum sein, da es ne nach oben geöffnete parabel ist
Subtract untill you can't multiply! :B[/i]
what school/grade is this? :D
i have never been amazing in maths but this is just too easy XD
Hence, we differentiate f(x) = 5 * x^2 + x - 16:
f'(x) = 5 * 2 * x ^1 + x^0 - 0
f'(x) = 10 * x + 1
When f'(x) = 0:
0 = 10 * x + 1
10 * x = -1
x = -1 / 10
x = -0.1
So,
f(x) = 5 * (-0.1)^2 - 0.1 - 16
f(x) = 0.05 - 0.1 - 16
f(x) = -16.05
If you want to take it further to check it's a minima you can differentiate again. If f''(x) > 0, then it's a minima, if f''(x) < 0, it's a maxima, if f''(x)=0, it's a turning point.
f''(x) = 10 > 0
Hence, the stationary point at x = -0.1 is a minima.
But why is it the right way? I dont get it...
The first derivative of a function of x tells you the gradient (or slope) of that function at whatever value of x you put into it. When the gradient is zero, that means you're either right at the bottom of a 'valley' (in a minima), at the top of a peak (a maxima) or in an area where the curve goes through an s-shaped bend (a turning point). Since the derivative can be used to find when the gradient is zero, that's why it works to find the minima.
The second derivative of a function tells you in which direction the gradient is changing. If the gradient is increasing the second derivative will be positive - that means, at a stationary point where the gradient is zero, in order to increase it must be changing from a negative gradient to a positive gradient - if you think about that, that could only happen at a minima.
If you want a simpler solution, you'll be able to solve this pretty easily graphically, if you draw it out neatly. Do a graph of the function between, say x=-2 and x=2, with 0.1 increments and you should get the answer.
thanks for ur help.
next time i gonna know who to ask hehe :)
Can you even call it turning point when f'' (x) isn't 0 ?!
Without the Fourier transform you wouldn't hear any sound from your PC.
Without matrix/vector algebra you wouldn't see any graphics on your PC.
And what would engineering be without differential equations?
Scientists use high level maths a lot more, to derive the correlations and stuff that the engineers use, but the engineers don't touch that stuff very often.
That doesn't mean engineers don't learn advanced maths myself at uni - it's just not used very much.
f'(x) = 10x + 1 (extrema = 0, so x = -0.1)
Proof it's a minimum, not a maxium:
f''(x) = 10 (>0 so minimum)