Mixing
Ways to find $min[f(x)]$ where $f(x) = (x-1)(x-2)(x-3)(x-4)$ without using derivatives
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
As title states I need to:
Find $min[f(x)]$ where $f(x) = (x-1)(x-2)(x-3)(x-4)$ without using derivatives
Since I i'm restricted to not use the derivatives I've started to play with the function in different ways. After some experiments I've noticed the following:
Let
$$
g(x) = (x-1)(x-3)
$$
and
$$
h(x) = (x-2)(x-4)
$$
Then i tried to find vertexes with $x_v = -b over 2a$ and calculate values of $g(x)$ and $h(x)$ in $x_v$ points and they appear to be the minimum values for $f(x)$. I've also checked this for $p(x) = (x-1)(x-2)(x-5)(x-6)$ and a lot of other similar polynomials. All of them are symmetric with respect to some $x$.
Based on the above the $min[f(x)] = -1$ and $min[g(x)] = -4$ but I'm not sure why this worked. Could someone explain me what happened? I would
also appreciate if anyone could tell whether there exists a general way of finding minimum for even power polynomials of the following kind:
$$
prod_k=1^2n(x-k)
$$
algebra-precalculus polynomials maxima-minima
asked Jul 16 at 16:13
roman
4391413
add a comment |Â
up vote
3
down vote
favorite
As title states I need to:
Find $min[f(x)]$ where $f(x) = (x-1)(x-2)(x-3)(x-4)$ without using derivatives
Since I i'm restricted to not use the derivatives I've started to play with the function in different ways. After some experiments I've noticed the following:
Let
$$
g(x) = (x-1)(x-3)
$$
and
$$
h(x) = (x-2)(x-4)
$$
Then i tried to find vertexes with $x_v = -b over 2a$ and calculate values of $g(x)$ and $h(x)$ in $x_v$ points and they appear to be the minimum values for $f(x)$. I've also checked this for $p(x) = (x-1)(x-2)(x-5)(x-6)$ and a lot of other similar polynomials. All of them are symmetric with respect to some $x$.
Based on the above the $min[f(x)] = -1$ and $min[g(x)] = -4$ but I'm not sure why this worked. Could someone explain me what happened? I would
also appreciate if anyone could tell whether there exists a general way of finding minimum for even power polynomials of the following kind:
$$
prod_k=1^2n(x-k)
$$
algebra-precalculus polynomials maxima-minima
asked Jul 16 at 16:13
roman
4391413
5
Let $g(x)=f(x+5/2)$.
– Lord Shark the Unknown
Jul 16 at 16:14
1
math.stackexchange.com/questions/1295586/…
– Mythomorphic
Jul 16 at 16:37
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
As title states I need to:
Find $min[f(x)]$ where $f(x) = (x-1)(x-2)(x-3)(x-4)$ without using derivatives
Since I i'm restricted to not use the derivatives I've started to play with the function in different ways. After some experiments I've noticed the following:
Let
$$
g(x) = (x-1)(x-3)
$$
and
$$
h(x) = (x-2)(x-4)
$$
Then i tried to find vertexes with $x_v = -b over 2a$ and calculate values of $g(x)$ and $h(x)$ in $x_v$ points and they appear to be the minimum values for $f(x)$. I've also checked this for $p(x) = (x-1)(x-2)(x-5)(x-6)$ and a lot of other similar polynomials. All of them are symmetric with respect to some $x$.
Based on the above the $min[f(x)] = -1$ and $min[g(x)] = -4$ but I'm not sure why this worked. Could someone explain me what happened? I would
also appreciate if anyone could tell whether there exists a general way of finding minimum for even power polynomials of the following kind:
$$
prod_k=1^2n(x-k)
$$
algebra-precalculus polynomials maxima-minima
asked Jul 16 at 16:13
roman
4391413
As title states I need to:
Find $min[f(x)]$ where $f(x) = (x-1)(x-2)(x-3)(x-4)$ without using derivatives
Since I i'm restricted to not use the derivatives I've started to play with the function in different ways. After some experiments I've noticed the following:
Let
$$
g(x) = (x-1)(x-3)
$$
and
$$
h(x) = (x-2)(x-4)
$$
Then i tried to find vertexes with $x_v = -b over 2a$ and calculate values of $g(x)$ and $h(x)$ in $x_v$ points and they appear to be the minimum values for $f(x)$. I've also checked this for $p(x) = (x-1)(x-2)(x-5)(x-6)$ and a lot of other similar polynomials. All of them are symmetric with respect to some $x$.
Based on the above the $min[f(x)] = -1$ and $min[g(x)] = -4$ but I'm not sure why this worked. Could someone explain me what happened? I would
also appreciate if anyone could tell whether there exists a general way of finding minimum for even power polynomials of the following kind:
$$
prod_k=1^2n(x-k)
$$
algebra-precalculus polynomials maxima-minima
asked Jul 16 at 16:13
roman
4391413
edited Jul 17 at 10:00
asked Jul 16 at 16:13
roman
4391413
asked Jul 16 at 16:13
roman
4391413
asked Jul 16 at 16:13
roman
4391413
4391413
5
Let $g(x)=f(x+5/2)$.
– Lord Shark the Unknown
Jul 16 at 16:14
1
math.stackexchange.com/questions/1295586/…
– Mythomorphic
Jul 16 at 16:37
add a comment |Â
5
Let $g(x)=f(x+5/2)$.
– Lord Shark the Unknown
Jul 16 at 16:14
1
math.stackexchange.com/questions/1295586/…
– Mythomorphic
Jul 16 at 16:37
5
5
Let $g(x)=f(x+5/2)$.
– Lord Shark the Unknown
Jul 16 at 16:14
Let $g(x)=f(x+5/2)$.
– Lord Shark the Unknown
Jul 16 at 16:14
1
1
math.stackexchange.com/questions/1295586/…
– Mythomorphic
Jul 16 at 16:37
math.stackexchange.com/questions/1295586/…
– Mythomorphic
Jul 16 at 16:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
Hint: You can exploit symmetry. Note the zeroes are equally spaced, so their center is $$tfrac1+2+3+44=tfrac104=tfrac52$$
This means you can write
$$f(x)=(x-tfrac52-tfrac32)(x-tfrac52-tfrac12)(x-tfrac52+tfrac12)(x-tfrac52+tfrac32)$$
$$=left((x-tfrac52)^2-(tfrac32)^2right)left((x-tfrac52)^2-(tfrac12)^2right)$$
$$=(z-tfrac94)(z-tfrac14)$$
where $z=(x-tfrac52)^2$. Now you can find the $z_0 geq 0$ which minimizes this polynomial using symmetry (vertex of the parabola). Then $f(z_0)$ is the desired quantity.
answered Jul 16 at 16:35
MPW
28.5k11853
add a comment |Â
up vote
10
down vote
$$beginarrayrcl
(x-1)(x-2)(x-3)(x-4)
&=& (x^2-5x+4)(x^2-5x+6) \
&=& (x^2-5x+5)^2-1 \
&=& ((x-2.5)^2-1.25)^2-1 \
&ge& -1
endarray$$
Equality is achieved when $(x-2.5)^2 - 1.25 = 0$.
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Hint: You can exploit symmetry. Note the zeroes are equally spaced, so their center is $$tfrac1+2+3+44=tfrac104=tfrac52$$
This means you can write
$$f(x)=(x-tfrac52-tfrac32)(x-tfrac52-tfrac12)(x-tfrac52+tfrac12)(x-tfrac52+tfrac32)$$
$$=left((x-tfrac52)^2-(tfrac32)^2right)left((x-tfrac52)^2-(tfrac12)^2right)$$
$$=(z-tfrac94)(z-tfrac14)$$
where $z=(x-tfrac52)^2$. Now you can find the $z_0 geq 0$ which minimizes this polynomial using symmetry (vertex of the parabola). Then $f(z_0)$ is the desired quantity.
answered Jul 16 at 16:35
MPW
28.5k11853
add a comment |Â
up vote
9
down vote
accepted
Hint: You can exploit symmetry. Note the zeroes are equally spaced, so their center is $$tfrac1+2+3+44=tfrac104=tfrac52$$
This means you can write
$$f(x)=(x-tfrac52-tfrac32)(x-tfrac52-tfrac12)(x-tfrac52+tfrac12)(x-tfrac52+tfrac32)$$
$$=left((x-tfrac52)^2-(tfrac32)^2right)left((x-tfrac52)^2-(tfrac12)^2right)$$
$$=(z-tfrac94)(z-tfrac14)$$
where $z=(x-tfrac52)^2$. Now you can find the $z_0 geq 0$ which minimizes this polynomial using symmetry (vertex of the parabola). Then $f(z_0)$ is the desired quantity.
answered Jul 16 at 16:35
MPW
28.5k11853
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Hint: You can exploit symmetry. Note the zeroes are equally spaced, so their center is $$tfrac1+2+3+44=tfrac104=tfrac52$$
This means you can write
$$f(x)=(x-tfrac52-tfrac32)(x-tfrac52-tfrac12)(x-tfrac52+tfrac12)(x-tfrac52+tfrac32)$$
$$=left((x-tfrac52)^2-(tfrac32)^2right)left((x-tfrac52)^2-(tfrac12)^2right)$$
$$=(z-tfrac94)(z-tfrac14)$$
where $z=(x-tfrac52)^2$. Now you can find the $z_0 geq 0$ which minimizes this polynomial using symmetry (vertex of the parabola). Then $f(z_0)$ is the desired quantity.
answered Jul 16 at 16:35
MPW
28.5k11853
Hint: You can exploit symmetry. Note the zeroes are equally spaced, so their center is $$tfrac1+2+3+44=tfrac104=tfrac52$$
This means you can write
$$f(x)=(x-tfrac52-tfrac32)(x-tfrac52-tfrac12)(x-tfrac52+tfrac12)(x-tfrac52+tfrac32)$$
$$=left((x-tfrac52)^2-(tfrac32)^2right)left((x-tfrac52)^2-(tfrac12)^2right)$$
$$=(z-tfrac94)(z-tfrac14)$$
where $z=(x-tfrac52)^2$. Now you can find the $z_0 geq 0$ which minimizes this polynomial using symmetry (vertex of the parabola). Then $f(z_0)$ is the desired quantity.
answered Jul 16 at 16:35
MPW
28.5k11853
answered Jul 16 at 16:35
MPW
28.5k11853
answered Jul 16 at 16:35
MPW
28.5k11853
answered Jul 16 at 16:35
MPW
28.5k11853
28.5k11853
add a comment |Â
add a comment |Â
up vote
10
down vote
$$beginarrayrcl
(x-1)(x-2)(x-3)(x-4)
&=& (x^2-5x+4)(x^2-5x+6) \
&=& (x^2-5x+5)^2-1 \
&=& ((x-2.5)^2-1.25)^2-1 \
&ge& -1
endarray$$
Equality is achieved when $(x-2.5)^2 - 1.25 = 0$.
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
add a comment |Â
up vote
10
down vote
$$beginarrayrcl
(x-1)(x-2)(x-3)(x-4)
&=& (x^2-5x+4)(x^2-5x+6) \
&=& (x^2-5x+5)^2-1 \
&=& ((x-2.5)^2-1.25)^2-1 \
&ge& -1
endarray$$
Equality is achieved when $(x-2.5)^2 - 1.25 = 0$.
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
add a comment |Â
up vote
10
down vote
up vote
10
down vote
$$beginarrayrcl
(x-1)(x-2)(x-3)(x-4)
&=& (x^2-5x+4)(x^2-5x+6) \
&=& (x^2-5x+5)^2-1 \
&=& ((x-2.5)^2-1.25)^2-1 \
&ge& -1
endarray$$
Equality is achieved when $(x-2.5)^2 - 1.25 = 0$.
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
$$beginarrayrcl
(x-1)(x-2)(x-3)(x-4)
&=& (x^2-5x+4)(x^2-5x+6) \
&=& (x^2-5x+5)^2-1 \
&=& ((x-2.5)^2-1.25)^2-1 \
&ge& -1
endarray$$
Equality is achieved when $(x-2.5)^2 - 1.25 = 0$.
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
answered Jul 16 at 16:28
Kenny Lau
18.7k2157
18.7k2157
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853557%2fways-to-find-minfx-where-fx-x-1x-2x-3x-4-without-using-deri%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
5
Let $g(x)=f(x+5/2)$.
– Lord Shark the Unknown
Jul 16 at 16:14
1
math.stackexchange.com/questions/1295586/…
– Mythomorphic
Jul 16 at 16:37