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Find a value of $a$ for which line segment equals 4
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$f(x)=2^x+1+2^4-x$
There is a value of $a$ for which the line $y=a$ intersects the graph of $f$ in two points $P$ and $Q$. The length of the line segment $PQ$ equals $4$.
I tried to figure it out on my own. Looking at the graph I saw that $y=a$ would intercept the graph twice from $f$'s minima, which is at $x=3/2$. Yet I'm unsure how to make an equation that would give me the answer regarding the line segment.
Would appreciate help! Thanks in advance.
calculus algebra-precalculus
asked Jul 22 at 9:30
Eli S.
1199
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up vote
1
down vote
favorite
$f(x)=2^x+1+2^4-x$
There is a value of $a$ for which the line $y=a$ intersects the graph of $f$ in two points $P$ and $Q$. The length of the line segment $PQ$ equals $4$.
I tried to figure it out on my own. Looking at the graph I saw that $y=a$ would intercept the graph twice from $f$'s minima, which is at $x=3/2$. Yet I'm unsure how to make an equation that would give me the answer regarding the line segment.
Would appreciate help! Thanks in advance.
calculus algebra-precalculus
asked Jul 22 at 9:30
Eli S.
1199
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$f(x)=2^x+1+2^4-x$
There is a value of $a$ for which the line $y=a$ intersects the graph of $f$ in two points $P$ and $Q$. The length of the line segment $PQ$ equals $4$.
I tried to figure it out on my own. Looking at the graph I saw that $y=a$ would intercept the graph twice from $f$'s minima, which is at $x=3/2$. Yet I'm unsure how to make an equation that would give me the answer regarding the line segment.
Would appreciate help! Thanks in advance.
calculus algebra-precalculus
asked Jul 22 at 9:30
Eli S.
1199
$f(x)=2^x+1+2^4-x$
There is a value of $a$ for which the line $y=a$ intersects the graph of $f$ in two points $P$ and $Q$. The length of the line segment $PQ$ equals $4$.
I tried to figure it out on my own. Looking at the graph I saw that $y=a$ would intercept the graph twice from $f$'s minima, which is at $x=3/2$. Yet I'm unsure how to make an equation that would give me the answer regarding the line segment.
Would appreciate help! Thanks in advance.
calculus algebra-precalculus
asked Jul 22 at 9:30
Eli S.
1199
asked Jul 22 at 9:30
Eli S.
1199
asked Jul 22 at 9:30
Eli S.
1199
asked Jul 22 at 9:30
Eli S.
1199
1199
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
So, if for some $b$ we have $f(b)=a$ then we have $f(b+4)=a$ (or $f(b-4)=a)$) .
So $$2^b+1+2^4-b = 2^b+5+2^-b$$
so $$2^-b(2^4-1)= 2^b+1(2^4-1)implies 2^b+1= 2^-bimplies b=-1over 2$$
so $a = f(-1/2) = 17sqrt2$.
answered Jul 22 at 10:06
greedoid
26.2k93473
Thank you, you made a small mistake there on the final answer. It's $17sqrt2$.
– Eli S.
Jul 22 at 17:44
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$$f(3/2+x)=2^3/2+x+1+2^4-(3/2+x)=2^frac52+x+2^frac52-x$$
$$f(3/2-x)=2^3/2-x+1+2^4-(3/2-x)=2^frac52+x+2^frac52-x$$
That is we have $$f(3/2+x)=f(3/2-x)$$
The graph is symmetrical about $x=frac32$.
Hence the $x$ values for the two points are $frac32+2$ and $frac32-2$.
Having that we should be able to find the value of $a$.
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
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up vote
0
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Correct me if wrong.
Let $z:=2^x$, then
$2z+2^4z^-1= a$, or
$z^2 -(a/2)z +2^3=0.$
Note:
$z_1/z_2= 2^x_1-x_2= 2^4$.
$z_1×z_2 = 2^3$ (Vieta).
$z_1^2 = 2^7$; or
$z_1=2^3+1/2$; i.e
$x_1=3+1/2$, and $x_2=-1/2$.
$-a/2= -z_1-z_2$ (Vieta);
$a=2(2^3+1/2+2^-1/2)=$
$2^42^1/2 + 2^1/2=√2(17)$.
answered Jul 22 at 11:42
Peter Szilas
7,9252617
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
So, if for some $b$ we have $f(b)=a$ then we have $f(b+4)=a$ (or $f(b-4)=a)$) .
So $$2^b+1+2^4-b = 2^b+5+2^-b$$
so $$2^-b(2^4-1)= 2^b+1(2^4-1)implies 2^b+1= 2^-bimplies b=-1over 2$$
so $a = f(-1/2) = 17sqrt2$.
answered Jul 22 at 10:06
greedoid
26.2k93473
Thank you, you made a small mistake there on the final answer. It's $17sqrt2$.
– Eli S.
Jul 22 at 17:44
add a comment |Â
up vote
2
down vote
accepted
So, if for some $b$ we have $f(b)=a$ then we have $f(b+4)=a$ (or $f(b-4)=a)$) .
So $$2^b+1+2^4-b = 2^b+5+2^-b$$
so $$2^-b(2^4-1)= 2^b+1(2^4-1)implies 2^b+1= 2^-bimplies b=-1over 2$$
so $a = f(-1/2) = 17sqrt2$.
answered Jul 22 at 10:06
greedoid
26.2k93473
Thank you, you made a small mistake there on the final answer. It's $17sqrt2$.
– Eli S.
Jul 22 at 17:44
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
So, if for some $b$ we have $f(b)=a$ then we have $f(b+4)=a$ (or $f(b-4)=a)$) .
So $$2^b+1+2^4-b = 2^b+5+2^-b$$
so $$2^-b(2^4-1)= 2^b+1(2^4-1)implies 2^b+1= 2^-bimplies b=-1over 2$$
so $a = f(-1/2) = 17sqrt2$.
answered Jul 22 at 10:06
greedoid
26.2k93473
So, if for some $b$ we have $f(b)=a$ then we have $f(b+4)=a$ (or $f(b-4)=a)$) .
So $$2^b+1+2^4-b = 2^b+5+2^-b$$
so $$2^-b(2^4-1)= 2^b+1(2^4-1)implies 2^b+1= 2^-bimplies b=-1over 2$$
so $a = f(-1/2) = 17sqrt2$.
answered Jul 22 at 10:06
greedoid
26.2k93473
edited Jul 22 at 18:04
answered Jul 22 at 10:06
greedoid
26.2k93473
answered Jul 22 at 10:06
greedoid
26.2k93473
answered Jul 22 at 10:06
greedoid
26.2k93473
26.2k93473
Thank you, you made a small mistake there on the final answer. It's $17sqrt2$.
– Eli S.
Jul 22 at 17:44
add a comment |Â
Thank you, you made a small mistake there on the final answer. It's $17sqrt2$.
– Eli S.
Jul 22 at 17:44
Thank you, you made a small mistake there on the final answer. It's $17sqrt2$.
– Eli S.
Jul 22 at 17:44
Thank you, you made a small mistake there on the final answer. It's $17sqrt2$.
– Eli S.
Jul 22 at 17:44
add a comment |Â
up vote
0
down vote
$$f(3/2+x)=2^3/2+x+1+2^4-(3/2+x)=2^frac52+x+2^frac52-x$$
$$f(3/2-x)=2^3/2-x+1+2^4-(3/2-x)=2^frac52+x+2^frac52-x$$
That is we have $$f(3/2+x)=f(3/2-x)$$
The graph is symmetrical about $x=frac32$.
Hence the $x$ values for the two points are $frac32+2$ and $frac32-2$.
Having that we should be able to find the value of $a$.
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
add a comment |Â
up vote
0
down vote
$$f(3/2+x)=2^3/2+x+1+2^4-(3/2+x)=2^frac52+x+2^frac52-x$$
$$f(3/2-x)=2^3/2-x+1+2^4-(3/2-x)=2^frac52+x+2^frac52-x$$
That is we have $$f(3/2+x)=f(3/2-x)$$
The graph is symmetrical about $x=frac32$.
Hence the $x$ values for the two points are $frac32+2$ and $frac32-2$.
Having that we should be able to find the value of $a$.
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$f(3/2+x)=2^3/2+x+1+2^4-(3/2+x)=2^frac52+x+2^frac52-x$$
$$f(3/2-x)=2^3/2-x+1+2^4-(3/2-x)=2^frac52+x+2^frac52-x$$
That is we have $$f(3/2+x)=f(3/2-x)$$
The graph is symmetrical about $x=frac32$.
Hence the $x$ values for the two points are $frac32+2$ and $frac32-2$.
Having that we should be able to find the value of $a$.
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
$$f(3/2+x)=2^3/2+x+1+2^4-(3/2+x)=2^frac52+x+2^frac52-x$$
$$f(3/2-x)=2^3/2-x+1+2^4-(3/2-x)=2^frac52+x+2^frac52-x$$
That is we have $$f(3/2+x)=f(3/2-x)$$
The graph is symmetrical about $x=frac32$.
Hence the $x$ values for the two points are $frac32+2$ and $frac32-2$.
Having that we should be able to find the value of $a$.
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
answered Jul 22 at 9:44
Siong Thye Goh
77.6k134795
77.6k134795
add a comment |Â
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Let $z:=2^x$, then
$2z+2^4z^-1= a$, or
$z^2 -(a/2)z +2^3=0.$
Note:
$z_1/z_2= 2^x_1-x_2= 2^4$.
$z_1×z_2 = 2^3$ (Vieta).
$z_1^2 = 2^7$; or
$z_1=2^3+1/2$; i.e
$x_1=3+1/2$, and $x_2=-1/2$.
$-a/2= -z_1-z_2$ (Vieta);
$a=2(2^3+1/2+2^-1/2)=$
$2^42^1/2 + 2^1/2=√2(17)$.
answered Jul 22 at 11:42
Peter Szilas
7,9252617
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Let $z:=2^x$, then
$2z+2^4z^-1= a$, or
$z^2 -(a/2)z +2^3=0.$
Note:
$z_1/z_2= 2^x_1-x_2= 2^4$.
$z_1×z_2 = 2^3$ (Vieta).
$z_1^2 = 2^7$; or
$z_1=2^3+1/2$; i.e
$x_1=3+1/2$, and $x_2=-1/2$.
$-a/2= -z_1-z_2$ (Vieta);
$a=2(2^3+1/2+2^-1/2)=$
$2^42^1/2 + 2^1/2=√2(17)$.
answered Jul 22 at 11:42
Peter Szilas
7,9252617
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Correct me if wrong.
Let $z:=2^x$, then
$2z+2^4z^-1= a$, or
$z^2 -(a/2)z +2^3=0.$
Note:
$z_1/z_2= 2^x_1-x_2= 2^4$.
$z_1×z_2 = 2^3$ (Vieta).
$z_1^2 = 2^7$; or
$z_1=2^3+1/2$; i.e
$x_1=3+1/2$, and $x_2=-1/2$.
$-a/2= -z_1-z_2$ (Vieta);
$a=2(2^3+1/2+2^-1/2)=$
$2^42^1/2 + 2^1/2=√2(17)$.
answered Jul 22 at 11:42
Peter Szilas
7,9252617
Correct me if wrong.
Let $z:=2^x$, then
$2z+2^4z^-1= a$, or
$z^2 -(a/2)z +2^3=0.$
Note:
$z_1/z_2= 2^x_1-x_2= 2^4$.
$z_1×z_2 = 2^3$ (Vieta).
$z_1^2 = 2^7$; or
$z_1=2^3+1/2$; i.e
$x_1=3+1/2$, and $x_2=-1/2$.
$-a/2= -z_1-z_2$ (Vieta);
$a=2(2^3+1/2+2^-1/2)=$
$2^42^1/2 + 2^1/2=√2(17)$.
answered Jul 22 at 11:42
Peter Szilas
7,9252617
answered Jul 22 at 11:42
Peter Szilas
7,9252617
answered Jul 22 at 11:42
Peter Szilas
7,9252617
answered Jul 22 at 11:42
Peter Szilas
7,9252617
7,9252617
add a comment |Â
add a comment |Â
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