(Calculus) Derivative Thinking Question
$begingroup$
Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.
The Question:
Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?
Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.
Find the required conditions on the parameters a,b,c in terms of a.
My Progress So Far:
So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.
Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.
I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.
calculus derivatives quadratics tangent-line slope
edited 3 hours ago
Jon due
948
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.
The Question:
Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?
Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.
Find the required conditions on the parameters a,b,c in terms of a.
My Progress So Far:
So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.
Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.
I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.
calculus derivatives quadratics tangent-line slope
edited 3 hours ago
Jon due
948
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
3 hours ago
$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
3 hours ago
1
$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
3 hours ago
$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
3 hours ago
$begingroup$
If we needed $f'(x)=0$ for all $x$, then we would require $a=b=0$. However, you don't need $f'(x)=0$ for all $x$, you just want there to exist at least one $x$ for which $f'(x)=0$. Assuming $a$ and $b$ are any values that satsify $aneq 0$, the answer below (which I upvoted when it came out) shows that this works for $x^* = -b/(2a)$. Now if $a=0$, we require $b=0$, which makes sense since then the function and its reciprocal are just constants. Another subtle issue is that you need to ensure $f(-b/(2a))neq 0$, else the answer does not apply as the reciprocal is not defined there.
$endgroup$
– Michael
1 hour ago
|
show 1 more comment
$begingroup$
Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.
The Question:
Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?
Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.
Find the required conditions on the parameters a,b,c in terms of a.
My Progress So Far:
So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.
Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.
I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.
calculus derivatives quadratics tangent-line slope
edited 3 hours ago
Jon due
948
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.
The Question:
Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?
Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.
Find the required conditions on the parameters a,b,c in terms of a.
My Progress So Far:
So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.
Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.
I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.
calculus derivatives quadratics tangent-line slope
calculus derivatives quadratics tangent-line slope
edited 3 hours ago
Jon due
948
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Jon due
948
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Jon due
948
edited 3 hours ago
Jon due
948
edited 3 hours ago
Jon due
948
948
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
asked 4 hours ago
Rasheed AmanzaiRasheed Amanzai
243
243
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
3 hours ago
$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
3 hours ago
1
$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
3 hours ago
$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
3 hours ago
$begingroup$
If we needed $f'(x)=0$ for all $x$, then we would require $a=b=0$. However, you don't need $f'(x)=0$ for all $x$, you just want there to exist at least one $x$ for which $f'(x)=0$. Assuming $a$ and $b$ are any values that satsify $aneq 0$, the answer below (which I upvoted when it came out) shows that this works for $x^* = -b/(2a)$. Now if $a=0$, we require $b=0$, which makes sense since then the function and its reciprocal are just constants. Another subtle issue is that you need to ensure $f(-b/(2a))neq 0$, else the answer does not apply as the reciprocal is not defined there.
$endgroup$
– Michael
1 hour ago
|
show 1 more comment
$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
3 hours ago
$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
3 hours ago
1
$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
3 hours ago
$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
3 hours ago
$begingroup$
If we needed $f'(x)=0$ for all $x$, then we would require $a=b=0$. However, you don't need $f'(x)=0$ for all $x$, you just want there to exist at least one $x$ for which $f'(x)=0$. Assuming $a$ and $b$ are any values that satsify $aneq 0$, the answer below (which I upvoted when it came out) shows that this works for $x^* = -b/(2a)$. Now if $a=0$, we require $b=0$, which makes sense since then the function and its reciprocal are just constants. Another subtle issue is that you need to ensure $f(-b/(2a))neq 0$, else the answer does not apply as the reciprocal is not defined there.
$endgroup$
– Michael
1 hour ago
$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
3 hours ago
$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
3 hours ago
$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
3 hours ago
$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
3 hours ago
1
1
$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
3 hours ago
$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
3 hours ago
$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
3 hours ago
$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
3 hours ago
$begingroup$
If we needed $f'(x)=0$ for all $x$, then we would require $a=b=0$. However, you don't need $f'(x)=0$ for all $x$, you just want there to exist at least one $x$ for which $f'(x)=0$. Assuming $a$ and $b$ are any values that satsify $aneq 0$, the answer below (which I upvoted when it came out) shows that this works for $x^* = -b/(2a)$. Now if $a=0$, we require $b=0$, which makes sense since then the function and its reciprocal are just constants. Another subtle issue is that you need to ensure $f(-b/(2a))neq 0$, else the answer does not apply as the reciprocal is not defined there.
$endgroup$
– Michael
1 hour ago
$begingroup$
If we needed $f'(x)=0$ for all $x$, then we would require $a=b=0$. However, you don't need $f'(x)=0$ for all $x$, you just want there to exist at least one $x$ for which $f'(x)=0$. Assuming $a$ and $b$ are any values that satsify $aneq 0$, the answer below (which I upvoted when it came out) shows that this works for $x^* = -b/(2a)$. Now if $a=0$, we require $b=0$, which makes sense since then the function and its reciprocal are just constants. Another subtle issue is that you need to ensure $f(-b/(2a))neq 0$, else the answer does not apply as the reciprocal is not defined there.
$endgroup$
– Michael
1 hour ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If $g(x)=1/f(x)$, then, by the chain rule,
$$
g'(x)=-frac{f'(x)}{f(x)^2}
$$
Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
$$
-frac{f'(x)}{f(x)^2}=f'(x)
$$
which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.
You can observe that this is independent of $f$ being a quadratic polynomial.
In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.
answered 3 hours ago
egregegreg
184k1486205
$endgroup$
add a comment |
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$begingroup$
If $g(x)=1/f(x)$, then, by the chain rule,
$$
g'(x)=-frac{f'(x)}{f(x)^2}
$$
Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
$$
-frac{f'(x)}{f(x)^2}=f'(x)
$$
which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.
You can observe that this is independent of $f$ being a quadratic polynomial.
In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.
answered 3 hours ago
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
If $g(x)=1/f(x)$, then, by the chain rule,
$$
g'(x)=-frac{f'(x)}{f(x)^2}
$$
Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
$$
-frac{f'(x)}{f(x)^2}=f'(x)
$$
which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.
You can observe that this is independent of $f$ being a quadratic polynomial.
In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.
answered 3 hours ago
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
If $g(x)=1/f(x)$, then, by the chain rule,
$$
g'(x)=-frac{f'(x)}{f(x)^2}
$$
Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
$$
-frac{f'(x)}{f(x)^2}=f'(x)
$$
which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.
You can observe that this is independent of $f$ being a quadratic polynomial.
In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.
answered 3 hours ago
egregegreg
184k1486205
$endgroup$
If $g(x)=1/f(x)$, then, by the chain rule,
$$
g'(x)=-frac{f'(x)}{f(x)^2}
$$
Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
$$
-frac{f'(x)}{f(x)^2}=f'(x)
$$
which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.
You can observe that this is independent of $f$ being a quadratic polynomial.
In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.
answered 3 hours ago
egregegreg
184k1486205
answered 3 hours ago
egregegreg
184k1486205
answered 3 hours ago
egregegreg
184k1486205
answered 3 hours ago
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
Rasheed Amanzai is a new contributor. Be nice, and check out our Code of Conduct.
Rasheed Amanzai is a new contributor. Be nice, and check out our Code of Conduct.
Rasheed Amanzai is a new contributor. Be nice, and check out our Code of Conduct.
Rasheed Amanzai is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
3 hours ago
$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
3 hours ago
1
$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
3 hours ago
$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
3 hours ago
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If we needed $f'(x)=0$ for all $x$, then we would require $a=b=0$. However, you don't need $f'(x)=0$ for all $x$, you just want there to exist at least one $x$ for which $f'(x)=0$. Assuming $a$ and $b$ are any values that satsify $aneq 0$, the answer below (which I upvoted when it came out) shows that this works for $x^* = -b/(2a)$. Now if $a=0$, we require $b=0$, which makes sense since then the function and its reciprocal are just constants. Another subtle issue is that you need to ensure $f(-b/(2a))neq 0$, else the answer does not apply as the reciprocal is not defined there.
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– Michael
1 hour ago