A pre-calculus problem about a quadratic functionIs this a problem that has already been solved?How to know...
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A pre-calculus problem about a quadratic function
Is this a problem that has already been solved?How to know when a line is parallel to the xz-planeFind range of $sin x+cos^{2}x$Proof that the graph of a linear function and its inverse cannot be perpendicular.Function exercise check-upIs the graph of $(cos x)(sec x)$ discontinuous?Is there an error in this AP Calculus quiz on function transformations?If $f$ continuous and $[f(x_1)+…+f(x_n)]/n=f([x_1+…+x_n]/n)$ then $f$ linearConfusing high school questionProve that every even degree polynomial function has a maximum or minimum in $mathbb{R}$
$begingroup$
This is from a high school junior math test (I'm not a high school student); The problem as designed to take less than 20 minutes, preferably 10-15. Judging from its source, this is not a trick question, and I am pretty sure all information given must be used.
Also, we cannot use continuity, sequence, differentiation or integration.
Let $f(x)$ be a quadratic polynomial with a positive leading coefficient. Let $g(x) = 1 - frac{2}{x-5}$ with the domain $x < 5$. For any real number $t < 3$, let $h(t)$ be the minimum of $f(g(x))$ for $t leq x leq t+2$. Suppose also that $h(t)= f(g(t+2))$ when $t < 1$ and $h(t) = 6$ for $1 leq t < 3$ and that $h(-1) =7$. Find $f(5)$.
It suffices to find an explicit formula for $f$, which is the direction I am heading at.
Here is my work so far: using the fact that the leading coefficient of $f$ is positive and that $g$ is strixtly increasing, I can deduce that the "tip" of the graph of $f$ lies on or to the right of the line $x=3$. Also, $h(-1) =7$ is exactly the statement that $f(frac32) = 7$.
My intuition tells me that the tip must lie on the line $x=3$, but I am not sure how to prove it. The difficulty lies in the fact that, even if the tip of $f$ lies not on the line, I am not exactly sure that such would imply $f$ wouldn't be constant at some subinterval of $1 leq t < 3$. I am slightly unsure of how to use the fact that $h(t) = 6$ on $[1,3]$. If my first conjecture is true, then $f(g(x)), t leq x leq t+2$ with fixed $1 leq t < 3$, must have its minimum when $f$ has its mimimum so long as $g^{-1}(3)=4$ lies on the interval... am I right? Please correct me if I am wrong.
EDIT: by playing around with Desmos, I may be wrong on the claim that the tip of $f$ must lie on or to the right of $x=3$...
Any help would be greatly appreciated.
functions problem-solving
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
$endgroup$
add a comment |
$begingroup$
This is from a high school junior math test (I'm not a high school student); The problem as designed to take less than 20 minutes, preferably 10-15. Judging from its source, this is not a trick question, and I am pretty sure all information given must be used.
Also, we cannot use continuity, sequence, differentiation or integration.
Let $f(x)$ be a quadratic polynomial with a positive leading coefficient. Let $g(x) = 1 - frac{2}{x-5}$ with the domain $x < 5$. For any real number $t < 3$, let $h(t)$ be the minimum of $f(g(x))$ for $t leq x leq t+2$. Suppose also that $h(t)= f(g(t+2))$ when $t < 1$ and $h(t) = 6$ for $1 leq t < 3$ and that $h(-1) =7$. Find $f(5)$.
It suffices to find an explicit formula for $f$, which is the direction I am heading at.
Here is my work so far: using the fact that the leading coefficient of $f$ is positive and that $g$ is strixtly increasing, I can deduce that the "tip" of the graph of $f$ lies on or to the right of the line $x=3$. Also, $h(-1) =7$ is exactly the statement that $f(frac32) = 7$.
My intuition tells me that the tip must lie on the line $x=3$, but I am not sure how to prove it. The difficulty lies in the fact that, even if the tip of $f$ lies not on the line, I am not exactly sure that such would imply $f$ wouldn't be constant at some subinterval of $1 leq t < 3$. I am slightly unsure of how to use the fact that $h(t) = 6$ on $[1,3]$. If my first conjecture is true, then $f(g(x)), t leq x leq t+2$ with fixed $1 leq t < 3$, must have its minimum when $f$ has its mimimum so long as $g^{-1}(3)=4$ lies on the interval... am I right? Please correct me if I am wrong.
EDIT: by playing around with Desmos, I may be wrong on the claim that the tip of $f$ must lie on or to the right of $x=3$...
Any help would be greatly appreciated.
functions problem-solving
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
$endgroup$
$begingroup$
Tip of $f(g(x))$ lies on $x=3$, so tip of $f$ lies on $x = g(3) = 1 - 2/(-2) = 2$.
$endgroup$
– enedil
Mar 9 at 20:12
$begingroup$
@enedil Thank you. Why is it that the tip of $f(g(x))$ lies on $x=3$? I sense that you are using the fact that $h$ is constant on $[1,3)$ but could you prove it rigorously? Fyi, the fact that $f$ is strictly decreasing doesn't mean that $f(g(x))$ is not constant, e.g. the interval $[0,1]$ with $f=x^2$, $g=1/x^2$.
$endgroup$
– Cute Brownie
Mar 10 at 2:35
add a comment |
$begingroup$
This is from a high school junior math test (I'm not a high school student); The problem as designed to take less than 20 minutes, preferably 10-15. Judging from its source, this is not a trick question, and I am pretty sure all information given must be used.
Also, we cannot use continuity, sequence, differentiation or integration.
Let $f(x)$ be a quadratic polynomial with a positive leading coefficient. Let $g(x) = 1 - frac{2}{x-5}$ with the domain $x < 5$. For any real number $t < 3$, let $h(t)$ be the minimum of $f(g(x))$ for $t leq x leq t+2$. Suppose also that $h(t)= f(g(t+2))$ when $t < 1$ and $h(t) = 6$ for $1 leq t < 3$ and that $h(-1) =7$. Find $f(5)$.
It suffices to find an explicit formula for $f$, which is the direction I am heading at.
Here is my work so far: using the fact that the leading coefficient of $f$ is positive and that $g$ is strixtly increasing, I can deduce that the "tip" of the graph of $f$ lies on or to the right of the line $x=3$. Also, $h(-1) =7$ is exactly the statement that $f(frac32) = 7$.
My intuition tells me that the tip must lie on the line $x=3$, but I am not sure how to prove it. The difficulty lies in the fact that, even if the tip of $f$ lies not on the line, I am not exactly sure that such would imply $f$ wouldn't be constant at some subinterval of $1 leq t < 3$. I am slightly unsure of how to use the fact that $h(t) = 6$ on $[1,3]$. If my first conjecture is true, then $f(g(x)), t leq x leq t+2$ with fixed $1 leq t < 3$, must have its minimum when $f$ has its mimimum so long as $g^{-1}(3)=4$ lies on the interval... am I right? Please correct me if I am wrong.
EDIT: by playing around with Desmos, I may be wrong on the claim that the tip of $f$ must lie on or to the right of $x=3$...
Any help would be greatly appreciated.
functions problem-solving
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
$endgroup$
This is from a high school junior math test (I'm not a high school student); The problem as designed to take less than 20 minutes, preferably 10-15. Judging from its source, this is not a trick question, and I am pretty sure all information given must be used.
Also, we cannot use continuity, sequence, differentiation or integration.
Let $f(x)$ be a quadratic polynomial with a positive leading coefficient. Let $g(x) = 1 - frac{2}{x-5}$ with the domain $x < 5$. For any real number $t < 3$, let $h(t)$ be the minimum of $f(g(x))$ for $t leq x leq t+2$. Suppose also that $h(t)= f(g(t+2))$ when $t < 1$ and $h(t) = 6$ for $1 leq t < 3$ and that $h(-1) =7$. Find $f(5)$.
It suffices to find an explicit formula for $f$, which is the direction I am heading at.
Here is my work so far: using the fact that the leading coefficient of $f$ is positive and that $g$ is strixtly increasing, I can deduce that the "tip" of the graph of $f$ lies on or to the right of the line $x=3$. Also, $h(-1) =7$ is exactly the statement that $f(frac32) = 7$.
My intuition tells me that the tip must lie on the line $x=3$, but I am not sure how to prove it. The difficulty lies in the fact that, even if the tip of $f$ lies not on the line, I am not exactly sure that such would imply $f$ wouldn't be constant at some subinterval of $1 leq t < 3$. I am slightly unsure of how to use the fact that $h(t) = 6$ on $[1,3]$. If my first conjecture is true, then $f(g(x)), t leq x leq t+2$ with fixed $1 leq t < 3$, must have its minimum when $f$ has its mimimum so long as $g^{-1}(3)=4$ lies on the interval... am I right? Please correct me if I am wrong.
EDIT: by playing around with Desmos, I may be wrong on the claim that the tip of $f$ must lie on or to the right of $x=3$...
Any help would be greatly appreciated.
functions problem-solving
functions problem-solving
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
edited Mar 10 at 3:42
Cute Brownie
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
asked Mar 9 at 17:54
Cute BrownieCute Brownie
1,043417
1,043417
$begingroup$
Tip of $f(g(x))$ lies on $x=3$, so tip of $f$ lies on $x = g(3) = 1 - 2/(-2) = 2$.
$endgroup$
– enedil
Mar 9 at 20:12
$begingroup$
@enedil Thank you. Why is it that the tip of $f(g(x))$ lies on $x=3$? I sense that you are using the fact that $h$ is constant on $[1,3)$ but could you prove it rigorously? Fyi, the fact that $f$ is strictly decreasing doesn't mean that $f(g(x))$ is not constant, e.g. the interval $[0,1]$ with $f=x^2$, $g=1/x^2$.
$endgroup$
– Cute Brownie
Mar 10 at 2:35
add a comment |
$begingroup$
Tip of $f(g(x))$ lies on $x=3$, so tip of $f$ lies on $x = g(3) = 1 - 2/(-2) = 2$.
$endgroup$
– enedil
Mar 9 at 20:12
$begingroup$
@enedil Thank you. Why is it that the tip of $f(g(x))$ lies on $x=3$? I sense that you are using the fact that $h$ is constant on $[1,3)$ but could you prove it rigorously? Fyi, the fact that $f$ is strictly decreasing doesn't mean that $f(g(x))$ is not constant, e.g. the interval $[0,1]$ with $f=x^2$, $g=1/x^2$.
$endgroup$
– Cute Brownie
Mar 10 at 2:35
$begingroup$
Tip of $f(g(x))$ lies on $x=3$, so tip of $f$ lies on $x = g(3) = 1 - 2/(-2) = 2$.
$endgroup$
– enedil
Mar 9 at 20:12
$begingroup$
Tip of $f(g(x))$ lies on $x=3$, so tip of $f$ lies on $x = g(3) = 1 - 2/(-2) = 2$.
$endgroup$
– enedil
Mar 9 at 20:12
$begingroup$
@enedil Thank you. Why is it that the tip of $f(g(x))$ lies on $x=3$? I sense that you are using the fact that $h$ is constant on $[1,3)$ but could you prove it rigorously? Fyi, the fact that $f$ is strictly decreasing doesn't mean that $f(g(x))$ is not constant, e.g. the interval $[0,1]$ with $f=x^2$, $g=1/x^2$.
$endgroup$
– Cute Brownie
Mar 10 at 2:35
$begingroup$
@enedil Thank you. Why is it that the tip of $f(g(x))$ lies on $x=3$? I sense that you are using the fact that $h$ is constant on $[1,3)$ but could you prove it rigorously? Fyi, the fact that $f$ is strictly decreasing doesn't mean that $f(g(x))$ is not constant, e.g. the interval $[0,1]$ with $f=x^2$, $g=1/x^2$.
$endgroup$
– Cute Brownie
Mar 10 at 2:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As $g$ is increasing on whole domain (and it's an infinite interval), composition $f(g(x))$ increases where $f(x)$ increases. There is the "tip", say $g(x_0)$ s.t. for all smaller numbers, $f$ is decreasing and for all bigger, $f$ is increasing (basic properties of quadratics, and uses the fact that first coefficient of $f$ is positive). That means that if $x_0>t+2$, we know that $f(g(x))$ is strictly decreasing on $[t, t+2]$, therefore minimum is at right end of the interval. If $t leq x_0 leq t+2$, then minimum of $f(g(x))$ is at $x_0$.
This in particular means that $x_0=3$, so the lowest point of $f$ is at $g(3)$.
So you have $7=h(-1)=f(g(1))=f(3/2)$ and $f(g(3))=6$, $-b/a=g(3)$, where $f(x)=ax^2+bx+c$.
answered Mar 10 at 2:45
enedilenedil
1,466620
$endgroup$
$begingroup$
Thank you. Could you elaborate on your statement "if $t leq x leq t+2$, then minimum of $f(g(x))$ is at $x_0$?" I am not sure if this is the case, since it may be the case that $g(x_0)$ will give us the point $x_1$ which will make $f$ itself minimum on the interval.
$endgroup$
– Cute Brownie
Mar 10 at 3:36
$begingroup$
@CuteBrownie you're right, I was writing it being half asleep. I Will correct this difference between notation.
$endgroup$
– enedil
Mar 10 at 11:13
add a comment |
Your Answer
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$begingroup$
As $g$ is increasing on whole domain (and it's an infinite interval), composition $f(g(x))$ increases where $f(x)$ increases. There is the "tip", say $g(x_0)$ s.t. for all smaller numbers, $f$ is decreasing and for all bigger, $f$ is increasing (basic properties of quadratics, and uses the fact that first coefficient of $f$ is positive). That means that if $x_0>t+2$, we know that $f(g(x))$ is strictly decreasing on $[t, t+2]$, therefore minimum is at right end of the interval. If $t leq x_0 leq t+2$, then minimum of $f(g(x))$ is at $x_0$.
This in particular means that $x_0=3$, so the lowest point of $f$ is at $g(3)$.
So you have $7=h(-1)=f(g(1))=f(3/2)$ and $f(g(3))=6$, $-b/a=g(3)$, where $f(x)=ax^2+bx+c$.
answered Mar 10 at 2:45
enedilenedil
1,466620
$endgroup$
$begingroup$
Thank you. Could you elaborate on your statement "if $t leq x leq t+2$, then minimum of $f(g(x))$ is at $x_0$?" I am not sure if this is the case, since it may be the case that $g(x_0)$ will give us the point $x_1$ which will make $f$ itself minimum on the interval.
$endgroup$
– Cute Brownie
Mar 10 at 3:36
$begingroup$
@CuteBrownie you're right, I was writing it being half asleep. I Will correct this difference between notation.
$endgroup$
– enedil
Mar 10 at 11:13
add a comment |
$begingroup$
As $g$ is increasing on whole domain (and it's an infinite interval), composition $f(g(x))$ increases where $f(x)$ increases. There is the "tip", say $g(x_0)$ s.t. for all smaller numbers, $f$ is decreasing and for all bigger, $f$ is increasing (basic properties of quadratics, and uses the fact that first coefficient of $f$ is positive). That means that if $x_0>t+2$, we know that $f(g(x))$ is strictly decreasing on $[t, t+2]$, therefore minimum is at right end of the interval. If $t leq x_0 leq t+2$, then minimum of $f(g(x))$ is at $x_0$.
This in particular means that $x_0=3$, so the lowest point of $f$ is at $g(3)$.
So you have $7=h(-1)=f(g(1))=f(3/2)$ and $f(g(3))=6$, $-b/a=g(3)$, where $f(x)=ax^2+bx+c$.
answered Mar 10 at 2:45
enedilenedil
1,466620
$endgroup$
$begingroup$
Thank you. Could you elaborate on your statement "if $t leq x leq t+2$, then minimum of $f(g(x))$ is at $x_0$?" I am not sure if this is the case, since it may be the case that $g(x_0)$ will give us the point $x_1$ which will make $f$ itself minimum on the interval.
$endgroup$
– Cute Brownie
Mar 10 at 3:36
$begingroup$
@CuteBrownie you're right, I was writing it being half asleep. I Will correct this difference between notation.
$endgroup$
– enedil
Mar 10 at 11:13
add a comment |
$begingroup$
As $g$ is increasing on whole domain (and it's an infinite interval), composition $f(g(x))$ increases where $f(x)$ increases. There is the "tip", say $g(x_0)$ s.t. for all smaller numbers, $f$ is decreasing and for all bigger, $f$ is increasing (basic properties of quadratics, and uses the fact that first coefficient of $f$ is positive). That means that if $x_0>t+2$, we know that $f(g(x))$ is strictly decreasing on $[t, t+2]$, therefore minimum is at right end of the interval. If $t leq x_0 leq t+2$, then minimum of $f(g(x))$ is at $x_0$.
This in particular means that $x_0=3$, so the lowest point of $f$ is at $g(3)$.
So you have $7=h(-1)=f(g(1))=f(3/2)$ and $f(g(3))=6$, $-b/a=g(3)$, where $f(x)=ax^2+bx+c$.
answered Mar 10 at 2:45
enedilenedil
1,466620
$endgroup$
As $g$ is increasing on whole domain (and it's an infinite interval), composition $f(g(x))$ increases where $f(x)$ increases. There is the "tip", say $g(x_0)$ s.t. for all smaller numbers, $f$ is decreasing and for all bigger, $f$ is increasing (basic properties of quadratics, and uses the fact that first coefficient of $f$ is positive). That means that if $x_0>t+2$, we know that $f(g(x))$ is strictly decreasing on $[t, t+2]$, therefore minimum is at right end of the interval. If $t leq x_0 leq t+2$, then minimum of $f(g(x))$ is at $x_0$.
This in particular means that $x_0=3$, so the lowest point of $f$ is at $g(3)$.
So you have $7=h(-1)=f(g(1))=f(3/2)$ and $f(g(3))=6$, $-b/a=g(3)$, where $f(x)=ax^2+bx+c$.
answered Mar 10 at 2:45
enedilenedil
1,466620
edited Mar 10 at 11:19
answered Mar 10 at 2:45
enedilenedil
1,466620
answered Mar 10 at 2:45
enedilenedil
1,466620
answered Mar 10 at 2:45
enedilenedil
1,466620
1,466620
$begingroup$
Thank you. Could you elaborate on your statement "if $t leq x leq t+2$, then minimum of $f(g(x))$ is at $x_0$?" I am not sure if this is the case, since it may be the case that $g(x_0)$ will give us the point $x_1$ which will make $f$ itself minimum on the interval.
$endgroup$
– Cute Brownie
Mar 10 at 3:36
$begingroup$
@CuteBrownie you're right, I was writing it being half asleep. I Will correct this difference between notation.
$endgroup$
– enedil
Mar 10 at 11:13
add a comment |
$begingroup$
Thank you. Could you elaborate on your statement "if $t leq x leq t+2$, then minimum of $f(g(x))$ is at $x_0$?" I am not sure if this is the case, since it may be the case that $g(x_0)$ will give us the point $x_1$ which will make $f$ itself minimum on the interval.
$endgroup$
– Cute Brownie
Mar 10 at 3:36
$begingroup$
@CuteBrownie you're right, I was writing it being half asleep. I Will correct this difference between notation.
$endgroup$
– enedil
Mar 10 at 11:13
$begingroup$
Thank you. Could you elaborate on your statement "if $t leq x leq t+2$, then minimum of $f(g(x))$ is at $x_0$?" I am not sure if this is the case, since it may be the case that $g(x_0)$ will give us the point $x_1$ which will make $f$ itself minimum on the interval.
$endgroup$
– Cute Brownie
Mar 10 at 3:36
$begingroup$
Thank you. Could you elaborate on your statement "if $t leq x leq t+2$, then minimum of $f(g(x))$ is at $x_0$?" I am not sure if this is the case, since it may be the case that $g(x_0)$ will give us the point $x_1$ which will make $f$ itself minimum on the interval.
$endgroup$
– Cute Brownie
Mar 10 at 3:36
$begingroup$
@CuteBrownie you're right, I was writing it being half asleep. I Will correct this difference between notation.
$endgroup$
– enedil
Mar 10 at 11:13
$begingroup$
@CuteBrownie you're right, I was writing it being half asleep. I Will correct this difference between notation.
$endgroup$
– enedil
Mar 10 at 11:13
add a comment |
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Tip of $f(g(x))$ lies on $x=3$, so tip of $f$ lies on $x = g(3) = 1 - 2/(-2) = 2$.
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– enedil
Mar 9 at 20:12
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@enedil Thank you. Why is it that the tip of $f(g(x))$ lies on $x=3$? I sense that you are using the fact that $h$ is constant on $[1,3)$ but could you prove it rigorously? Fyi, the fact that $f$ is strictly decreasing doesn't mean that $f(g(x))$ is not constant, e.g. the interval $[0,1]$ with $f=x^2$, $g=1/x^2$.
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– Cute Brownie
Mar 10 at 2:35