toandhsp

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9 years, 111 days

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These are answers submitted by toandhsp

g(x) = 3*x^2+2*m*x+2*m^2-3*m-6, x belongs to interval (1, ∞), -m/3 <=1.

We have g'(x) = 6x + 2m, g'(x) = 0 when and only when x = -m/3. Consider sign of g'(x),  We have g is increasing on (-m/3, +infinity). Note that 1>= -m/3. Therefore g'(x) >=0, for every x belongs to interval (1, ∞). So g in increasing in (1, ∞). Another way, g is a continuous function on [1, + ∞)), therefore g is  increasing in [1, ∞). 

We want to numer(df)>=0, for evrery x belongs to interval [1, ∞).

We have numer(df) = 3*x^2+2*m*x+2*m^2-3*m-6.

Consider the function g(x) = 3*x^2+2*m*x+2*m^2-3*m-6, x belongs to interval (1, ∞), -m/3 <=1.

It is easy to see that, g is a creasing function on interval [1, ∞). Therefore, for any x >=1, we have g(x) >= g(1).

Thus, g(x) >=0, for evrery x belongs to interval [1, ∞) when and only when g(1) >=0. In my code, i denotes g(1) by B. That is why i did solve([B>=0, -m/3<=1],m);

I used the command minimize such

minimize(3*x^2+2*m*x+2*m^2-3*m-6,1 <= x)

But it can not. How can i find the least of the function g:=x>3*x^2+2*m*x+2*m^2-3*m-6,  here x >=1.

Please help me. Thank you.

This is my way

s3:= x <-2 or x>-1;

s4:= x<=-3 or x>=3;

solve(evalb(s3 or s4));

But I do not know how to change from s1 to s3 and from s2 to s4. Please help me. Thank you.

 

Ok. your ansewr is coincident my answer

restart;f:=x->(x^2 + 2*m*x  + m+2)/(3*x+m):

A:=simplify(diff(f(x),x)):

N:=numer(A):

B:=subs(x=1,N):

solve([B>=0, -m/3<=1],m);

we need to find m so that df := simplify(diff(f(x), x)) >=1 for evrery x belongs to interval (1, ∞). That is mean minimum(numer(df))>=0, for evrery x belongs to interval [1, ∞) and -m/3 <=1. I think so.

I think your answer is not right. For example, with m = 0. 

With m = 0, the given equation has the form f(x) = (x^2 + 2)/(3*x).

This function is not creasing in the interval (1; + infinity). I did

g:=x->1/3*(x^2+2)/x;

solve(diff(g(x),x)>0);

We must to find all values of m which df(x) >=0, for every x belongs to inteval (1, +infty).

First cases. The function f(x) is increasing for x < -m/3) and for x>-m/3 if and only if the numerator of df is nonnegative. 

Secend cases. When discriminant := discrim(numer(df), x) >0, Suppose x_1, x_2 are different solutions of numer(df), x_1<x_2. We must find m so that x_2 >=1 and -m/3<=1.

From two cases, I think, our anserw are -3<=m<=1 or m >=3/2.

This is my code (i don't sure it is right).

restart;f:=x->(x^2 + 2*m*x  + m+2)/(3*x+m):

A:=simplify(diff(f(x),x)):

N:=numer(A):

B:=subs(x=1,N):

solve([B>=0, -m/3<=1],m);

This is my explain. Please comment to me. Thank you.

When subtitute m = -1 or m = 2, ... into the given function, we get a function is also increasing in (1; +infinity).

When subtitute m = 0 into the given function, we have f(x) = (x^2+2)/(3*x). This fuction is also increasing in (1; +infinity).

 f:=x->x*(2011+sqrt(2013-x^2));

 maximize(f(x),x=-sqrt(2013)..sqrt(2013),location);

 minimize(f(x),x=-sqrt(2013)..sqrt(2013),location);



Thank you.

If i only want to solve the given equation on the set of all Real numbers, what must i do?

Thank you very much.

The first is right. I am sorry. 

Problem. write the equation of the line cut the two lines d1: x = 2*t, y= -t+1, z = t-2, d2: x = -1+2*m, y = 1+m, z = 3 and perpendicular to the plane 7*x+y-4*z=0. 

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