Find the Local Maxima of f(x)=2x^3-3x^2-12x+3

[chjopl]

Find the local maxima:
f(x)=2x^3 - 3x^2 -12x + 3

 

[Planckenstein]

Find the local maxima:
f(x)=2x^3 - 3x^2 -12x + 3

Taking the derivative with respect to x, we get 6x^2 - 6x - 12. Simplified to 6(x^2 - x + 2). That there's a parabola facing up, so it doesn't have a local max, but does have an absolute min.

 

[chjopl]

does a local max exist for f(x)=x + ln(x)

 

[Planckenstein]

does a local max exist for f(x)=x + ln(x)

Same deal. The derivative is 1 + 1/x, which is 1/x shifted up one unit. Goes off into infinity, as well.

Unless these have a bounded domain, they both don't have local maxes.

 

[franznietzsche]

Same deal. The derivative is 1 + 1/x, which is 1/x shifted up one unit. Goes off into infinity, as well.

Unless these have a bounded domain, they both don't have local maxes.

THere is a local extremum for that, at x = -1. And it is a maximum (had to stop and think about the second derivative for a second.).

 

[franznietzsche]

Taking the derivative with respect to x, we get 6x^2 - 6x - 12. Simplified to 6(x^2 - x + 2). That there's a parabola facing up, so it doesn't have a local max, but does have an absolute min.

that should be 6(x^2 - x - 2) = 6(x+1)(x-2)

And the original function does have both a min and a max, just look at its graph.

the fact that the derivative has two zeros means it has to have one of each, you can't have only two of one.

Also the second derivative is not constant, so the two extremums have to be different. (one a min, the other a max)

 

[kreil]

Just because the derivative has 2 zeroes doesn't mean the function has to have a max and a min...it could pass through the x-axis at 1 point (the zero) and come back up, never becoming negative.

In order for their to be a max or min at a certain point, the zeroes of the first derivative are the places it would happen.

Next, to be sure, you need to make sure that either the derivative goes from + to - or - to + through that zero, or that the second derivative is positive/negative at that point (same thing, different methods). It is entirely possible you will make a sign chart and notice that the derivative is positive or negative on both sides...if so, it is not a local max or min.

For the question...f(x)=2x^3 - 3x^2 -12x + 3

f'(x)=6x^2-6x-12

6(x+1)(x-2)=0

x=-1, x=2

Sign Chart for f'(x):
at x=-2, f' is +
at x=-1 f' is 0
at x=0 f' is -
at x=2 f' is 0
at x=3 f' is +

f'(x) goes from + to - through x=-1, so x=-1 is a local max
f'(x) goes from - to + through x=2, so x=2 is a local min

 

[franznietzsche]

Just because the derivative has 2 zeroes doesn't mean the function has to have a max and a min...it could pass through the x-axis at 1 point (the zero) and come back up, never becoming negative.

In order for their to be a max or min at a certain point, the zeroes of the first derivative are the places it would happen.

That makes no sense.

You've contradicted yourself. If the first derivative has two zeros, there are two extremum. For every distinct, non-repeated factor in the derivative, there is an extremum. For every fact that appears more than once there can be one or zero extremum.

Since that there derivative had two distinct, non repeated factors, the oringial function had two extremum. AS noth i showed, and then you repeated.

 

[kreil]

What I said does not convey the idea I was trying to convey

Imagine the graph of a first derivative...it is positive and decreasing, touches the x-axis at 1 point (does NOT pass through the x axis), then increases. It's values over this interval are never negative. The graph of f changed concavity but does NOT have a local extremum at this point since the first derivative did not change signs.

That is what I was trying to say. Not every zero of the first derivative is guaranteed to be an extremum.

 

[franznietzsche]

What I said does not convey the idea I was trying to convey

Imagine the graph of a first derivative...it is positive and decreasing, touches the x-axis at 1 point (does NOT pass through the x axis), then increases. It's values over this interval are never negative. The graph of f changed concavity but does NOT have a local extremum at this point since the first derivative did not change signs.

That is what I was trying to say. Not every zero of the first derivative is guaranteed to be an extremum.

That only happens if the zero is from a repeated factor. the only way the sign does not change for a zero is if that factor is of the form (x-a)^2n, where n is some integer. Other wise, moving from one side of a to the other it changes signs.

So as I said: For a zero produced by any non-repeated factor in the first derivative, there is an extremum.

 

[dextercioby]

Until any of u 2 comes up with the mathematically viable proof of he's saying,lemme give u the following definitions.
1.Let f(x) be a continuous function upon it's entire domain.Then the point P is said to be a CRITICAL POINT TO f(x),if:f'(P)=0.

Solutions of the the eq.f'(x)=0 are said to be f(x)'s critical points.

2.Under the assumption above,the point P is said to be EXTREMUM POINT TO f(x),if:f'(P)=0 AND f''(P) is different from 0.

Thus,a function's extremum points are among its critical points.

WHEN TRYING TO GRAPH A FUNCTION,ALWAYS COMPUTE BOTH ITS FIRST AND SECOND DERIVATIVES.It's the only way to make sure that u're not missing an extremum,or maybe that a critical point is also an inflexion point,as it's the case for the simple f(x)=x^3.

 

[franznietzsche]

OK, its not that freakin complicated.


take
f ' (x) = (x-a)(x-b)^{2n}(x-c)^{3n}
where n is some integer such that n>0, and none of a, b, c are equal to each each other.

Obviously it has zeros at x= a, b, c.

then
f '' (x) = (x-b)^{2n}(x-c)^{3n} + 2(x-a)(x-b)^{2n-1}(x-c)^{3n} + 3(x-a)(x-b)^{2n}(x-c)^{3n-1}

so we get :

<br /> f &#039;&#039; (a) = (a-b)^{2n}(a-c)^{3n} + 0 + 0<br />

<br /> f &#039;&#039; (b) = 0 + 0 +0<br />

<br /> f &#039;&#039; (c) = 0 + 0 + 0<br />


So, dextercioby, as I was saying, for any non repeated factor, in a polynomial first derivative we have an extremum. Not that bloody complicated.