# Minima, maxima of function

by Дьявол
Tags: function, maxima, minima
 P: 365 How will I find minima,maxima or both of one function (if they exist)? For example: $$f(x)=x^2-3x+2$$; $$f(x)=x^2-2x-x+2$$; $$f(x)=x(x-2)-(x-2)$$; $$f(x)=(x-2)(x-1)$$ x1,2=1,2 Those are the zeroes of the function. But how will I find the critical points (minima,maxima) of the function if $x \in \mathbb{R}$
P: 362
 Quote by Дьявол How will I find minima,maxima or both of one function (if they exist)?
The minima/maxima of a function is the point(s) where $$f'(x)=0$$
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,691 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 x-intercepts of the function, you can use the fact that, for a parabola with vertical axis, the vertex always lies half way between the x-intercepts. Since this problem has nothing to do with linear algebra or abstract algebra, I am moving it to "General Mathematics".
 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 x-values) into your original function, and that will be your Y-coordinates. That will give you the values of your max's and min's. It will also let you see absolute max and min.
P: 810
 Quote by phyzmatix The minima/maxima of a function is the point(s) where $$f'(x)=0$$
P: 365
 Quote by Riogho Take the derivative of the originial function, then find where the derivative equals zero. Then plug those points (they are x-values) into your original function, and that will be your Y-coordinates. That will give you the values of your max's and min's. It will also let you see absolute max and min.
Do you mean like f(x)=(x-a)2+b ?

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
 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.
P: 810
 Quote by Desh627 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.

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?
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P: 39,691
 Quote by Дьявол How will I find minima,maxima or both of one function (if they exist)? For example: $$f(x)=x^2-3x+2$$;
 Quote by Дьявол Do you mean like f(x)=(x-a)2+b ? 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
??? It certainly is a parabola! Any quadratic function has a parabola as its graph. And the points (3,2), (2,0), (1,0), (-1,6), (-2,12), (-3,20) definitely lie on a parabola. The vertex is, as I said before, half way between 1 and 2, at 3/2. f(3/2)= 9/4- 9/2+ 2= -9/2+ 2= -5/2. f(x)= (x- 3/2)^2- 5/2.
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P: 15,204
 Quote by Tac-Tics 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?

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.
P: 125
 Quote by Tac-Tics 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?
Inflections points are the points where the 2nd derivative equals zero, at those points the concavity will change.

Where the first derivative is zero are called critical points, those are the relative max's and mins of a function.
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P: 15,204
 Quote by Riogho Where the first derivative is zero are called critical points
More appropriately, the points where the first derivative is zero are called stationary points. The absolute value function has a well-defined critical point: x=0. It has no stationary point because the derivative is non-zero everywhere except for x=0 and is not defined at x=0.
 those are the relative max's and mins of a function.
Not all stationary points are extremal. Tac-Tics gave a perfectly good example of a function that has a stationary point but no extremal points.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,691 A quadratic function does not have any "inflection" points. And the original question has already been completely answered.
P: 810
 Quote by Riogho Where the first derivative is zero are called critical points
Ah, yes. Thank you. It's been about 6 years since I studied this. The concept is there, but the terminology is fuzzy and needs dusting :-)
 P: 906 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).
 P: 365 Thanks for the posts. I found the minima with derivation of the original formula. So my goal is $x^2-3x+2=(x-a)^2+b$ $$x^2-3x+2=x^2-2ax+a^2+b$$ $$-3x+2=-2ax+(a^2+b)$$ $$-3=-2a$$ and $$2=(a^2+b)$$ So, a= 3/2 and $$b=2-9/4$$ $$b=-1/4$$ $$x^2-3x+2=(x-3/2)^2-1/4 \geq -1/4=f(3/2)$$ So 3/2 is the local minimum.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,691 In other words, the vertex of the parabola is at x= 3/2, half way between the x-intercepts. That's what I said in post 3.
 P: 365 I am again stuck with minima and maxima of function. The original problem: $$f(x)=\frac{4x}{3x^2+4}$$ Now, I tried: $$f(x)=\frac{4}{3}*\frac{x}{x^2+\frac{4}{3}}=\frac{4}{3}*\frac{1}{x+\frac {4}{3x}}$$ 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 $$\frac{4x}{3x^2+4} + 1 - 1=\frac{3(x^2+\frac{4}{3}x+\frac{4}{3})}{3(x^2+\frac{4}{3})} - 1 = \frac{(x-\frac{2}{3})^2+\frac{6}{9} \geq \frac{6}{9}}{(x-0)^2 + \frac{4}{3} \geq \frac{4}{3} } - 1$$ Again nothing. Please help me! Thanks in advance.

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