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Minima, maxima of function 
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#1
Oct3008, 04:30 AM

P: 365

How will I find minima,maxima or both of one function (if they exist)? For example:
[tex]f(x)=x^23x+2[/tex]; [tex]f(x)=x^22xx+2[/tex]; [tex]f(x)=x(x2)(x2)[/tex]; [tex]f(x)=(x2)(x1)[/tex] x_{1,2}=1,2 Those are the zeroes of the function. But how will I find the critical points (minima,maxima) of the function if [itex]x \in \mathbb{R}[/itex] 


#2
Oct3008, 06:31 AM

P: 362




#3
Oct3008, 07:28 AM

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PF Gold
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That's a quadratic function. Since it is concave upward it has a minimum but no maxiumum. The minimum is at the vertex you need to find the vertex. You could do that by completing the square but since you have already found the xintercepts of the function, you can use the fact that, for a parabola with vertical axis, the vertex always lies half way between the xintercepts.
Since this problem has nothing to do with linear algebra or abstract algebra, I am moving it to "General Mathematics". 


#4
Oct3008, 07:50 AM

P: 125

Minima, maxima of function
Take the derivative of the originial function, then find where the derivative equals zero. Then plug those points (they are xvalues) into your original function, and that will be your Ycoordinates. That will give you the values of your max's and min's. It will also let you see absolute max and min.



#5
Oct3008, 09:26 AM

P: 810




#6
Oct3008, 09:54 AM

P: 365

HallsofIvy, When I draw graphic of the function it is not parabola. f(3)=2 f(2)=0 f(1)=0 f(0)=2 f(1)=6 f(2)=12 f(3)=20 


#7
Oct3008, 10:53 AM

P: 40

f(x) has a relative minimum or maximum at either the endpoints of a region you're observing or when f'(x)=0. To determine whether it's a minimum, find when f''(x) is positive or negative; when f''(x) is positive, the graph is concave up, and when f''(x) is negative, the graph is concave down. Therefore, you have a relative extrema when f'(x)=0; it's a maximum if f''(x) is positive, or a minimum if f''(x) is negative.
F''(x)=0 are your points of inflection. 


#8
Oct3008, 11:07 AM

P: 810

Eh? If inflection points are where f''(x) = 0 (which they are), what do you call points where f'(x) = 0, but is neither a maximum nor a minimum? For example, f(x) = x^3, which has no maximum or minimum has f'(0) = 0. What are those points called if it's not inflection points? 


#9
Oct3008, 12:09 PM

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#10
Oct3008, 12:21 PM

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Edit Perhaps that was too hasty. The problem is that the term inflection point describes a point where the second derivative reaches this. Inflection points obviously do not in general coincide with stationary points. A saddle point is by definition a stationary point that is not a local extremum. 


#11
Oct3008, 12:25 PM

P: 125

Where the first derivative is zero are called critical points, those are the relative max's and mins of a function. 


#12
Oct3008, 12:34 PM

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#13
Oct3008, 02:21 PM

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A quadratic function does not have any "inflection" points. And the original question has already been completely answered.



#14
Oct3008, 03:08 PM

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#15
Oct3008, 08:23 PM

P: 898

Since all of your functions are quadratic, an alternative to calculus is to complete the square. Ie., your first function is f(x) = x^2  3x + 2 = (x  3/2)^2  5/2. Since squares of real numbers are never negative, your function has the minimum point (3/2, 5/2).



#16
Oct3108, 01:12 AM

P: 365

Thanks for the posts. I found the minima with derivation of the original formula. So my goal is [itex]x^23x+2=(xa)^2+b[/itex]
[tex]x^23x+2=x^22ax+a^2+b[/tex] [tex]3x+2=2ax+(a^2+b)[/tex] [tex]3=2a[/tex] and [tex]2=(a^2+b)[/tex] So, a= 3/2 and [tex]b=29/4[/tex] [tex]b=1/4[/tex] [tex] x^23x+2=(x3/2)^21/4 \geq 1/4=f(3/2)[/tex] So 3/2 is the local minimum. 


#17
Oct3108, 06:38 AM

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In other words, the vertex of the parabola is at x= 3/2, half way between the xintercepts. That's what I said in post 3.



#18
Dec708, 06:01 PM

P: 365

I am again stuck with minima and maxima of function.
The original problem: [tex]f(x)=\frac{4x}{3x^2+4}[/tex] Now, I tried: [tex]f(x)=\frac{4}{3}*\frac{x}{x^2+\frac{4}{3}}=\frac{4}{3}*\frac{1}{x+\frac {4}{3x}}[/tex] and I don't know how to go on out of here. I was trying to have something in the form (x+a)^{2}+b > or < b Also I tried [tex]\frac{4x}{3x^2+4} + 1  1=\frac{3(x^2+\frac{4}{3}x+\frac{4}{3})}{3(x^2+\frac{4}{3})}  1 = \frac{(x\frac{2}{3})^2+\frac{6}{9} \geq \frac{6}{9}}{(x0)^2 + \frac{4}{3} \geq \frac{4}{3} }  1[/tex] Again nothing. Please help me! Thanks in advance. 


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