What $L(x) = f(a) + f'(a)(x - a)$ has to do with this problem at all? I am confused? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The Line that is tangent to the curve?Calculus - Derivatives and Tangent LinesDerive equations of all lines which are tangent to the graph of y= -7 - x^2 and passing through the point (3,0). (This point is not on the graph).Finding Equation of tangent linefinding equation of tangent line, coming up with a half right answerhow to deduct the formula that gives the derivative of $f(x) = x^2$?I do not understand the alternative definition of derivative and how to use it to find tangent slope.When to use different formulas to find the slope of a tangent lineImplicit differentiation / find the equation of the tangent line using the derivativeWhy is the y-intercept of this calculus problem given like that in the solution?
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What $L(x) = f(a) + f'(a)(x - a)$ has to do with this problem at all? I am confused?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The Line that is tangent to the curve?Calculus - Derivatives and Tangent LinesDerive equations of all lines which are tangent to the graph of y= -7 - x^2 and passing through the point (3,0). (This point is not on the graph).Finding Equation of tangent linefinding equation of tangent line, coming up with a half right answerhow to deduct the formula that gives the derivative of $f(x) = x^2$?I do not understand the alternative definition of derivative and how to use it to find tangent slope.When to use different formulas to find the slope of a tangent lineImplicit differentiation / find the equation of the tangent line using the derivativeWhy is the y-intercept of this calculus problem given like that in the solution?
$begingroup$
Following is the word problem:
Given the formula $L(x) = f(a)+f'(a)(x-a)$, what is the linear approximation of the tangent line at $a=2$ for the following function $f(x)=x^3$?
After watching Khan Academy, I figured out that I have to find slope (derivative) of $f(x)$ which is $f'(x)=3x^2$, then plug-in 2 to get $f'(2)=3(2)^2=12$
then use point slope $(y-y1)=m(x-x1) implies (y$ is found by plugging in 2 in the original $f(x)=x^3)$ implies
$$(y-8)=12(x-2) implies y=12x-24+8 implies y=12x-16$$ and this is correct answer.
But I have no idea why problem mentions formula $L(x) = f(a)+f'(a)(x-a)$?
calculus derivatives
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
asked Apr 1 at 6:10
Aman KhanAman Khan
326
$endgroup$
add a comment |
$begingroup$
Following is the word problem:
Given the formula $L(x) = f(a)+f'(a)(x-a)$, what is the linear approximation of the tangent line at $a=2$ for the following function $f(x)=x^3$?
After watching Khan Academy, I figured out that I have to find slope (derivative) of $f(x)$ which is $f'(x)=3x^2$, then plug-in 2 to get $f'(2)=3(2)^2=12$
then use point slope $(y-y1)=m(x-x1) implies (y$ is found by plugging in 2 in the original $f(x)=x^3)$ implies
$$(y-8)=12(x-2) implies y=12x-24+8 implies y=12x-16$$ and this is correct answer.
But I have no idea why problem mentions formula $L(x) = f(a)+f'(a)(x-a)$?
calculus derivatives
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
asked Apr 1 at 6:10
Aman KhanAman Khan
326
$endgroup$
1
$begingroup$
It is the same thing you just did
$endgroup$
– randomgirl
Apr 1 at 6:14
6
$begingroup$
Compare $y-y_a =m(x-x_a )$ to $f(x)-f(a)=f'(a)(x-a)$ ... $f'(a)=m$ and some point on the line $(x_a ,y_a )=(a,f(a))$. They just solved for $f(x)$ and called it $L(x)$.
$endgroup$
– randomgirl
Apr 1 at 6:19
$begingroup$
Thanks randomgirl!
$endgroup$
– Aman Khan
Apr 1 at 6:22
add a comment |
$begingroup$
Following is the word problem:
Given the formula $L(x) = f(a)+f'(a)(x-a)$, what is the linear approximation of the tangent line at $a=2$ for the following function $f(x)=x^3$?
After watching Khan Academy, I figured out that I have to find slope (derivative) of $f(x)$ which is $f'(x)=3x^2$, then plug-in 2 to get $f'(2)=3(2)^2=12$
then use point slope $(y-y1)=m(x-x1) implies (y$ is found by plugging in 2 in the original $f(x)=x^3)$ implies
$$(y-8)=12(x-2) implies y=12x-24+8 implies y=12x-16$$ and this is correct answer.
But I have no idea why problem mentions formula $L(x) = f(a)+f'(a)(x-a)$?
calculus derivatives
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
asked Apr 1 at 6:10
Aman KhanAman Khan
326
$endgroup$
Following is the word problem:
Given the formula $L(x) = f(a)+f'(a)(x-a)$, what is the linear approximation of the tangent line at $a=2$ for the following function $f(x)=x^3$?
After watching Khan Academy, I figured out that I have to find slope (derivative) of $f(x)$ which is $f'(x)=3x^2$, then plug-in 2 to get $f'(2)=3(2)^2=12$
then use point slope $(y-y1)=m(x-x1) implies (y$ is found by plugging in 2 in the original $f(x)=x^3)$ implies
$$(y-8)=12(x-2) implies y=12x-24+8 implies y=12x-16$$ and this is correct answer.
But I have no idea why problem mentions formula $L(x) = f(a)+f'(a)(x-a)$?
calculus derivatives
calculus derivatives
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
asked Apr 1 at 6:10
Aman KhanAman Khan
326
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
asked Apr 1 at 6:10
Aman KhanAman Khan
326
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
edited Apr 1 at 6:23
Jimmy R.
33.3k42257
33.3k42257
asked Apr 1 at 6:10
Aman KhanAman Khan
326
asked Apr 1 at 6:10
Aman KhanAman Khan
326
asked Apr 1 at 6:10
Aman KhanAman Khan
326
326
1
$begingroup$
It is the same thing you just did
$endgroup$
– randomgirl
Apr 1 at 6:14
6
$begingroup$
Compare $y-y_a =m(x-x_a )$ to $f(x)-f(a)=f'(a)(x-a)$ ... $f'(a)=m$ and some point on the line $(x_a ,y_a )=(a,f(a))$. They just solved for $f(x)$ and called it $L(x)$.
$endgroup$
– randomgirl
Apr 1 at 6:19
$begingroup$
Thanks randomgirl!
$endgroup$
– Aman Khan
Apr 1 at 6:22
add a comment |
1
$begingroup$
It is the same thing you just did
$endgroup$
– randomgirl
Apr 1 at 6:14
6
$begingroup$
Compare $y-y_a =m(x-x_a )$ to $f(x)-f(a)=f'(a)(x-a)$ ... $f'(a)=m$ and some point on the line $(x_a ,y_a )=(a,f(a))$. They just solved for $f(x)$ and called it $L(x)$.
$endgroup$
– randomgirl
Apr 1 at 6:19
$begingroup$
Thanks randomgirl!
$endgroup$
– Aman Khan
Apr 1 at 6:22
1
1
$begingroup$
It is the same thing you just did
$endgroup$
– randomgirl
Apr 1 at 6:14
$begingroup$
It is the same thing you just did
$endgroup$
– randomgirl
Apr 1 at 6:14
6
6
$begingroup$
Compare $y-y_a =m(x-x_a )$ to $f(x)-f(a)=f'(a)(x-a)$ ... $f'(a)=m$ and some point on the line $(x_a ,y_a )=(a,f(a))$. They just solved for $f(x)$ and called it $L(x)$.
$endgroup$
– randomgirl
Apr 1 at 6:19
$begingroup$
Compare $y-y_a =m(x-x_a )$ to $f(x)-f(a)=f'(a)(x-a)$ ... $f'(a)=m$ and some point on the line $(x_a ,y_a )=(a,f(a))$. They just solved for $f(x)$ and called it $L(x)$.
$endgroup$
– randomgirl
Apr 1 at 6:19
$begingroup$
Thanks randomgirl!
$endgroup$
– Aman Khan
Apr 1 at 6:22
$begingroup$
Thanks randomgirl!
$endgroup$
– Aman Khan
Apr 1 at 6:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This L(x) is the general formula for the linear approximation that you are looking for. You just have to plug in your $f$. So, in your case $$L(x)=a^3+left.(x^3)'right|_a(x-a)=a^3+3a^2(x-a)=3a^2x-2a^3$$ which is linear (in $x$). Now, plugging in $a=2$ you can confirm your result $$L(x)=3(2^2)x-2(2^3)=12x-16$$
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
add a comment |
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$begingroup$
This L(x) is the general formula for the linear approximation that you are looking for. You just have to plug in your $f$. So, in your case $$L(x)=a^3+left.(x^3)'right|_a(x-a)=a^3+3a^2(x-a)=3a^2x-2a^3$$ which is linear (in $x$). Now, plugging in $a=2$ you can confirm your result $$L(x)=3(2^2)x-2(2^3)=12x-16$$
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
add a comment |
$begingroup$
This L(x) is the general formula for the linear approximation that you are looking for. You just have to plug in your $f$. So, in your case $$L(x)=a^3+left.(x^3)'right|_a(x-a)=a^3+3a^2(x-a)=3a^2x-2a^3$$ which is linear (in $x$). Now, plugging in $a=2$ you can confirm your result $$L(x)=3(2^2)x-2(2^3)=12x-16$$
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
add a comment |
$begingroup$
This L(x) is the general formula for the linear approximation that you are looking for. You just have to plug in your $f$. So, in your case $$L(x)=a^3+left.(x^3)'right|_a(x-a)=a^3+3a^2(x-a)=3a^2x-2a^3$$ which is linear (in $x$). Now, plugging in $a=2$ you can confirm your result $$L(x)=3(2^2)x-2(2^3)=12x-16$$
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
This L(x) is the general formula for the linear approximation that you are looking for. You just have to plug in your $f$. So, in your case $$L(x)=a^3+left.(x^3)'right|_a(x-a)=a^3+3a^2(x-a)=3a^2x-2a^3$$ which is linear (in $x$). Now, plugging in $a=2$ you can confirm your result $$L(x)=3(2^2)x-2(2^3)=12x-16$$
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
answered Apr 1 at 6:17
Jimmy R.Jimmy R.
33.3k42257
33.3k42257
add a comment |
add a comment |
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$begingroup$
It is the same thing you just did
$endgroup$
– randomgirl
Apr 1 at 6:14
6
$begingroup$
Compare $y-y_a =m(x-x_a )$ to $f(x)-f(a)=f'(a)(x-a)$ ... $f'(a)=m$ and some point on the line $(x_a ,y_a )=(a,f(a))$. They just solved for $f(x)$ and called it $L(x)$.
$endgroup$
– randomgirl
Apr 1 at 6:19
$begingroup$
Thanks randomgirl!
$endgroup$
– Aman Khan
Apr 1 at 6:22