How do I find ordered pair, given slope of the tangent line?Finding values of x that gives a specific slope...
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How do I find ordered pair, given slope of the tangent line?
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$begingroup$
The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 Rightarrow f^{prime}(x) = 3x^2 + 18x + 36 Rightarrow 3x^2 + 18x + 36 = 9 Rightarrow\\
& 3x^2 + 18x = -27 Rightarrow 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 Rightarrow
x = -3 x = -33
end{align*}
calculus functions derivatives tangent-line slope
edited Mar 10 at 5:16
Narasimham
20.9k62158
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 Rightarrow f^{prime}(x) = 3x^2 + 18x + 36 Rightarrow 3x^2 + 18x + 36 = 9 Rightarrow\\
& 3x^2 + 18x = -27 Rightarrow 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 Rightarrow
x = -3 x = -33
end{align*}
calculus functions derivatives tangent-line slope
edited Mar 10 at 5:16
Narasimham
20.9k62158
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
$ 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 $ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 3:55
1
$begingroup$
Which point are you given? If $x=9$ then you don't set the derivative equal to $9$, you plug $9$ into the derivative. Are you being asked to find the point on the curve whose derivative is $9$?
$endgroup$
– John Douma
Mar 10 at 4:02
$begingroup$
I meant $3x(x+6)=-27Rightarrow 3x=-27x+6$ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 4:03
$begingroup$
It's confusing if you use $x$ to represent the ordered pair, the independent variable, and the dependent variable
$endgroup$
– J. W. Tanner
Mar 10 at 4:16
$begingroup$
@JohnDouma: Since the answer is given to be $(-3,-44),$ I assume OP was asked to find the point on the curve where the derivative is $9$, though it's not stated clearly
$endgroup$
– J. W. Tanner
Mar 10 at 4:49
add a comment |
$begingroup$
The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 Rightarrow f^{prime}(x) = 3x^2 + 18x + 36 Rightarrow 3x^2 + 18x + 36 = 9 Rightarrow\\
& 3x^2 + 18x = -27 Rightarrow 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 Rightarrow
x = -3 x = -33
end{align*}
calculus functions derivatives tangent-line slope
edited Mar 10 at 5:16
Narasimham
20.9k62158
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 Rightarrow f^{prime}(x) = 3x^2 + 18x + 36 Rightarrow 3x^2 + 18x + 36 = 9 Rightarrow\\
& 3x^2 + 18x = -27 Rightarrow 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 Rightarrow
x = -3 x = -33
end{align*}
calculus functions derivatives tangent-line slope
calculus functions derivatives tangent-line slope
edited Mar 10 at 5:16
Narasimham
20.9k62158
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 5:16
Narasimham
20.9k62158
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 5:16
Narasimham
20.9k62158
edited Mar 10 at 5:16
Narasimham
20.9k62158
edited Mar 10 at 5:16
Narasimham
20.9k62158
20.9k62158
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
asked Mar 10 at 3:45
T. SaruwatariT. Saruwatari
61
61
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
T. Saruwatari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
$ 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 $ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 3:55
1
$begingroup$
Which point are you given? If $x=9$ then you don't set the derivative equal to $9$, you plug $9$ into the derivative. Are you being asked to find the point on the curve whose derivative is $9$?
$endgroup$
– John Douma
Mar 10 at 4:02
$begingroup$
I meant $3x(x+6)=-27Rightarrow 3x=-27x+6$ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 4:03
$begingroup$
It's confusing if you use $x$ to represent the ordered pair, the independent variable, and the dependent variable
$endgroup$
– J. W. Tanner
Mar 10 at 4:16
$begingroup$
@JohnDouma: Since the answer is given to be $(-3,-44),$ I assume OP was asked to find the point on the curve where the derivative is $9$, though it's not stated clearly
$endgroup$
– J. W. Tanner
Mar 10 at 4:49
add a comment |
$begingroup$
$ 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 $ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 3:55
1
$begingroup$
Which point are you given? If $x=9$ then you don't set the derivative equal to $9$, you plug $9$ into the derivative. Are you being asked to find the point on the curve whose derivative is $9$?
$endgroup$
– John Douma
Mar 10 at 4:02
$begingroup$
I meant $3x(x+6)=-27Rightarrow 3x=-27x+6$ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 4:03
$begingroup$
It's confusing if you use $x$ to represent the ordered pair, the independent variable, and the dependent variable
$endgroup$
– J. W. Tanner
Mar 10 at 4:16
$begingroup$
@JohnDouma: Since the answer is given to be $(-3,-44),$ I assume OP was asked to find the point on the curve where the derivative is $9$, though it's not stated clearly
$endgroup$
– J. W. Tanner
Mar 10 at 4:49
$begingroup$
$ 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 $ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 3:55
$begingroup$
$ 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 $ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 3:55
1
1
$begingroup$
Which point are you given? If $x=9$ then you don't set the derivative equal to $9$, you plug $9$ into the derivative. Are you being asked to find the point on the curve whose derivative is $9$?
$endgroup$
– John Douma
Mar 10 at 4:02
$begingroup$
Which point are you given? If $x=9$ then you don't set the derivative equal to $9$, you plug $9$ into the derivative. Are you being asked to find the point on the curve whose derivative is $9$?
$endgroup$
– John Douma
Mar 10 at 4:02
$begingroup$
I meant $3x(x+6)=-27Rightarrow 3x=-27x+6$ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 4:03
$begingroup$
I meant $3x(x+6)=-27Rightarrow 3x=-27x+6$ is not valid
$endgroup$
– J. W. Tanner
Mar 10 at 4:03
$begingroup$
It's confusing if you use $x$ to represent the ordered pair, the independent variable, and the dependent variable
$endgroup$
– J. W. Tanner
Mar 10 at 4:16
$begingroup$
It's confusing if you use $x$ to represent the ordered pair, the independent variable, and the dependent variable
$endgroup$
– J. W. Tanner
Mar 10 at 4:16
$begingroup$
@JohnDouma: Since the answer is given to be $(-3,-44),$ I assume OP was asked to find the point on the curve where the derivative is $9$, though it's not stated clearly
$endgroup$
– J. W. Tanner
Mar 10 at 4:49
$begingroup$
@JohnDouma: Since the answer is given to be $(-3,-44),$ I assume OP was asked to find the point on the curve where the derivative is $9$, though it's not stated clearly
$endgroup$
– J. W. Tanner
Mar 10 at 4:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You got $3x^2+18x=-27.$
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$ or $(x+3)^2=0$.
Can you take it from here?
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
$begingroup$
That's because I'm trying to find the slope of the tangent line at point 9, so I set it equal to 9 and 9 - 36 = -27
$endgroup$
– T. Saruwatari
Mar 10 at 3:57
$begingroup$
Also, I didn't get 3x=-27x+6=-27 those are separate, I got 3x = -27 AND x+6 = -27 in order to get an ordered pair
$endgroup$
– T. Saruwatari
Mar 10 at 3:59
$begingroup$
@T.Saruwatari: I agree with you that $3x^2+18x=-27$ follows from $f'(x)=9$; I was showing the correct solution from that point onward
$endgroup$
– J. W. Tanner
Mar 10 at 4:01
$begingroup$
That would only give me two sets of x = -3. But, I digress, I'm just going to take the L because I've spent 3 hours online searching for resources and I don't seem to be competent enough to figure it out
$endgroup$
– T. Saruwatari
Mar 10 at 4:16
$begingroup$
I thought you said the answer is $-3,$ and I showed how to get that: $x=-3$ and $f(x)=-44$
$endgroup$
– J. W. Tanner
Mar 10 at 4:18
|
show 1 more comment
$begingroup$
So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$.
First derive and set derivative to 9.
$ f'(x) = 3x^{2}+18x+36=9 $
$ Rightarrow x^{2}+6x+12-3=0 $ (divide by 3 then subtract 3)
$ Rightarrow (x+3)^{2}=0 $
$ therefore x=-3$ ($x$-ordinate of the point on f(x))
To find the $y$-oordinate substitute $x$ back into $f(x)$:
$f(-3)=-44$
Hence at the point $(-3,-44)$ the function $f(x)$ has a tangent slope gradient of $9$.
answered Mar 10 at 5:33
Daniel SDaniel S
33
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Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$frac{dy}{dx}= 3x^2+18x +36 $$
$$ frac{d^2y}{dx^2}=6x+18 $$
$$ frac{d^3y}{dx^3}=6 $$
$$ x=-3,y=-44,frac{dy}{dx}=9, frac{d^2y}{dx^2}=0 $$
all of which is seen in the WA plot
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
You got $3x^2+18x=-27.$
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$ or $(x+3)^2=0$.
Can you take it from here?
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
$begingroup$
That's because I'm trying to find the slope of the tangent line at point 9, so I set it equal to 9 and 9 - 36 = -27
$endgroup$
– T. Saruwatari
Mar 10 at 3:57
$begingroup$
Also, I didn't get 3x=-27x+6=-27 those are separate, I got 3x = -27 AND x+6 = -27 in order to get an ordered pair
$endgroup$
– T. Saruwatari
Mar 10 at 3:59
$begingroup$
@T.Saruwatari: I agree with you that $3x^2+18x=-27$ follows from $f'(x)=9$; I was showing the correct solution from that point onward
$endgroup$
– J. W. Tanner
Mar 10 at 4:01
$begingroup$
That would only give me two sets of x = -3. But, I digress, I'm just going to take the L because I've spent 3 hours online searching for resources and I don't seem to be competent enough to figure it out
$endgroup$
– T. Saruwatari
Mar 10 at 4:16
$begingroup$
I thought you said the answer is $-3,$ and I showed how to get that: $x=-3$ and $f(x)=-44$
$endgroup$
– J. W. Tanner
Mar 10 at 4:18
|
show 1 more comment
$begingroup$
You got $3x^2+18x=-27.$
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$ or $(x+3)^2=0$.
Can you take it from here?
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
$begingroup$
That's because I'm trying to find the slope of the tangent line at point 9, so I set it equal to 9 and 9 - 36 = -27
$endgroup$
– T. Saruwatari
Mar 10 at 3:57
$begingroup$
Also, I didn't get 3x=-27x+6=-27 those are separate, I got 3x = -27 AND x+6 = -27 in order to get an ordered pair
$endgroup$
– T. Saruwatari
Mar 10 at 3:59
$begingroup$
@T.Saruwatari: I agree with you that $3x^2+18x=-27$ follows from $f'(x)=9$; I was showing the correct solution from that point onward
$endgroup$
– J. W. Tanner
Mar 10 at 4:01
$begingroup$
That would only give me two sets of x = -3. But, I digress, I'm just going to take the L because I've spent 3 hours online searching for resources and I don't seem to be competent enough to figure it out
$endgroup$
– T. Saruwatari
Mar 10 at 4:16
$begingroup$
I thought you said the answer is $-3,$ and I showed how to get that: $x=-3$ and $f(x)=-44$
$endgroup$
– J. W. Tanner
Mar 10 at 4:18
|
show 1 more comment
$begingroup$
You got $3x^2+18x=-27.$
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$ or $(x+3)^2=0$.
Can you take it from here?
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
You got $3x^2+18x=-27.$
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$ or $(x+3)^2=0$.
Can you take it from here?
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
answered Mar 10 at 3:52
J. W. TannerJ. W. Tanner
3,2301320
3,2301320
$begingroup$
That's because I'm trying to find the slope of the tangent line at point 9, so I set it equal to 9 and 9 - 36 = -27
$endgroup$
– T. Saruwatari
Mar 10 at 3:57
$begingroup$
Also, I didn't get 3x=-27x+6=-27 those are separate, I got 3x = -27 AND x+6 = -27 in order to get an ordered pair
$endgroup$
– T. Saruwatari
Mar 10 at 3:59
$begingroup$
@T.Saruwatari: I agree with you that $3x^2+18x=-27$ follows from $f'(x)=9$; I was showing the correct solution from that point onward
$endgroup$
– J. W. Tanner
Mar 10 at 4:01
$begingroup$
That would only give me two sets of x = -3. But, I digress, I'm just going to take the L because I've spent 3 hours online searching for resources and I don't seem to be competent enough to figure it out
$endgroup$
– T. Saruwatari
Mar 10 at 4:16
$begingroup$
I thought you said the answer is $-3,$ and I showed how to get that: $x=-3$ and $f(x)=-44$
$endgroup$
– J. W. Tanner
Mar 10 at 4:18
|
show 1 more comment
$begingroup$
That's because I'm trying to find the slope of the tangent line at point 9, so I set it equal to 9 and 9 - 36 = -27
$endgroup$
– T. Saruwatari
Mar 10 at 3:57
$begingroup$
Also, I didn't get 3x=-27x+6=-27 those are separate, I got 3x = -27 AND x+6 = -27 in order to get an ordered pair
$endgroup$
– T. Saruwatari
Mar 10 at 3:59
$begingroup$
@T.Saruwatari: I agree with you that $3x^2+18x=-27$ follows from $f'(x)=9$; I was showing the correct solution from that point onward
$endgroup$
– J. W. Tanner
Mar 10 at 4:01
$begingroup$
That would only give me two sets of x = -3. But, I digress, I'm just going to take the L because I've spent 3 hours online searching for resources and I don't seem to be competent enough to figure it out
$endgroup$
– T. Saruwatari
Mar 10 at 4:16
$begingroup$
I thought you said the answer is $-3,$ and I showed how to get that: $x=-3$ and $f(x)=-44$
$endgroup$
– J. W. Tanner
Mar 10 at 4:18
$begingroup$
That's because I'm trying to find the slope of the tangent line at point 9, so I set it equal to 9 and 9 - 36 = -27
$endgroup$
– T. Saruwatari
Mar 10 at 3:57
$begingroup$
That's because I'm trying to find the slope of the tangent line at point 9, so I set it equal to 9 and 9 - 36 = -27
$endgroup$
– T. Saruwatari
Mar 10 at 3:57
$begingroup$
Also, I didn't get 3x=-27x+6=-27 those are separate, I got 3x = -27 AND x+6 = -27 in order to get an ordered pair
$endgroup$
– T. Saruwatari
Mar 10 at 3:59
$begingroup$
Also, I didn't get 3x=-27x+6=-27 those are separate, I got 3x = -27 AND x+6 = -27 in order to get an ordered pair
$endgroup$
– T. Saruwatari
Mar 10 at 3:59
$begingroup$
@T.Saruwatari: I agree with you that $3x^2+18x=-27$ follows from $f'(x)=9$; I was showing the correct solution from that point onward
$endgroup$
– J. W. Tanner
Mar 10 at 4:01
$begingroup$
@T.Saruwatari: I agree with you that $3x^2+18x=-27$ follows from $f'(x)=9$; I was showing the correct solution from that point onward
$endgroup$
– J. W. Tanner
Mar 10 at 4:01
$begingroup$
That would only give me two sets of x = -3. But, I digress, I'm just going to take the L because I've spent 3 hours online searching for resources and I don't seem to be competent enough to figure it out
$endgroup$
– T. Saruwatari
Mar 10 at 4:16
$begingroup$
That would only give me two sets of x = -3. But, I digress, I'm just going to take the L because I've spent 3 hours online searching for resources and I don't seem to be competent enough to figure it out
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– T. Saruwatari
Mar 10 at 4:16
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I thought you said the answer is $-3,$ and I showed how to get that: $x=-3$ and $f(x)=-44$
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– J. W. Tanner
Mar 10 at 4:18
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I thought you said the answer is $-3,$ and I showed how to get that: $x=-3$ and $f(x)=-44$
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– J. W. Tanner
Mar 10 at 4:18
|
show 1 more comment
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So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$.
First derive and set derivative to 9.
$ f'(x) = 3x^{2}+18x+36=9 $
$ Rightarrow x^{2}+6x+12-3=0 $ (divide by 3 then subtract 3)
$ Rightarrow (x+3)^{2}=0 $
$ therefore x=-3$ ($x$-ordinate of the point on f(x))
To find the $y$-oordinate substitute $x$ back into $f(x)$:
$f(-3)=-44$
Hence at the point $(-3,-44)$ the function $f(x)$ has a tangent slope gradient of $9$.
answered Mar 10 at 5:33
Daniel SDaniel S
33
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add a comment |
$begingroup$
So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$.
First derive and set derivative to 9.
$ f'(x) = 3x^{2}+18x+36=9 $
$ Rightarrow x^{2}+6x+12-3=0 $ (divide by 3 then subtract 3)
$ Rightarrow (x+3)^{2}=0 $
$ therefore x=-3$ ($x$-ordinate of the point on f(x))
To find the $y$-oordinate substitute $x$ back into $f(x)$:
$f(-3)=-44$
Hence at the point $(-3,-44)$ the function $f(x)$ has a tangent slope gradient of $9$.
answered Mar 10 at 5:33
Daniel SDaniel S
33
New contributor
Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$.
First derive and set derivative to 9.
$ f'(x) = 3x^{2}+18x+36=9 $
$ Rightarrow x^{2}+6x+12-3=0 $ (divide by 3 then subtract 3)
$ Rightarrow (x+3)^{2}=0 $
$ therefore x=-3$ ($x$-ordinate of the point on f(x))
To find the $y$-oordinate substitute $x$ back into $f(x)$:
$f(-3)=-44$
Hence at the point $(-3,-44)$ the function $f(x)$ has a tangent slope gradient of $9$.
answered Mar 10 at 5:33
Daniel SDaniel S
33
New contributor
Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$.
First derive and set derivative to 9.
$ f'(x) = 3x^{2}+18x+36=9 $
$ Rightarrow x^{2}+6x+12-3=0 $ (divide by 3 then subtract 3)
$ Rightarrow (x+3)^{2}=0 $
$ therefore x=-3$ ($x$-ordinate of the point on f(x))
To find the $y$-oordinate substitute $x$ back into $f(x)$:
$f(-3)=-44$
Hence at the point $(-3,-44)$ the function $f(x)$ has a tangent slope gradient of $9$.
answered Mar 10 at 5:33
Daniel SDaniel S
33
New contributor
Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 10 at 5:33
Daniel SDaniel S
33
New contributor
Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 10 at 5:33
Daniel SDaniel S
33
answered Mar 10 at 5:33
Daniel SDaniel S
33
33
New contributor
Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Daniel S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$frac{dy}{dx}= 3x^2+18x +36 $$
$$ frac{d^2y}{dx^2}=6x+18 $$
$$ frac{d^3y}{dx^3}=6 $$
$$ x=-3,y=-44,frac{dy}{dx}=9, frac{d^2y}{dx^2}=0 $$
all of which is seen in the WA plot
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$frac{dy}{dx}= 3x^2+18x +36 $$
$$ frac{d^2y}{dx^2}=6x+18 $$
$$ frac{d^3y}{dx^3}=6 $$
$$ x=-3,y=-44,frac{dy}{dx}=9, frac{d^2y}{dx^2}=0 $$
all of which is seen in the WA plot
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$frac{dy}{dx}= 3x^2+18x +36 $$
$$ frac{d^2y}{dx^2}=6x+18 $$
$$ frac{d^3y}{dx^3}=6 $$
$$ x=-3,y=-44,frac{dy}{dx}=9, frac{d^2y}{dx^2}=0 $$
all of which is seen in the WA plot
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
$endgroup$
From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$frac{dy}{dx}= 3x^2+18x +36 $$
$$ frac{d^2y}{dx^2}=6x+18 $$
$$ frac{d^3y}{dx^3}=6 $$
$$ x=-3,y=-44,frac{dy}{dx}=9, frac{d^2y}{dx^2}=0 $$
all of which is seen in the WA plot
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
edited Mar 10 at 6:04
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
answered Mar 10 at 5:11
NarasimhamNarasimham
20.9k62158
20.9k62158
add a comment |
add a comment |
T. Saruwatari is a new contributor. Be nice, and check out our Code of Conduct.
T. Saruwatari is a new contributor. Be nice, and check out our Code of Conduct.
T. Saruwatari is a new contributor. Be nice, and check out our Code of Conduct.
T. Saruwatari is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
$ 3x ( x + 6 ) = -27 Rightarrow 3x = -27 x + 6 = -27 $ is not valid
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– J. W. Tanner
Mar 10 at 3:55
1
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Which point are you given? If $x=9$ then you don't set the derivative equal to $9$, you plug $9$ into the derivative. Are you being asked to find the point on the curve whose derivative is $9$?
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– John Douma
Mar 10 at 4:02
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I meant $3x(x+6)=-27Rightarrow 3x=-27x+6$ is not valid
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– J. W. Tanner
Mar 10 at 4:03
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It's confusing if you use $x$ to represent the ordered pair, the independent variable, and the dependent variable
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– J. W. Tanner
Mar 10 at 4:16
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@JohnDouma: Since the answer is given to be $(-3,-44),$ I assume OP was asked to find the point on the curve where the derivative is $9$, though it's not stated clearly
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– J. W. Tanner
Mar 10 at 4:49