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Calculus: Finding an unknown coefficient in continuous equations
the word “derivative”Minimum speed using calculuscalculus integration, average height of point on semi circleCalculus, specifically deriving the rule for exponentsuniform probability density funcion area( continuous)Need advice in regards to CalculusIf $f$ is differntiable at $x_0$, then $f$ is continuous at $x_0$. How does this work?Average height of a point in half circleWhy do we seek for the $x$ where $f(x) = 0$?Solving for kinematics equations with calculus
$begingroup$
I am watching a video trying to understand continuity in calculus. The author first starts with the $ax - 4, x < 1$, and we are trying to find the value of "a".
The author subsequently derives $ax - 4$ as $7x - 4$, where $a = 7$.
Why did he come to a conclusion where $a$ is $7$, where $x < 1$?
calculus
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
$endgroup$
add a comment |
$begingroup$
I am watching a video trying to understand continuity in calculus. The author first starts with the $ax - 4, x < 1$, and we are trying to find the value of "a".
The author subsequently derives $ax - 4$ as $7x - 4$, where $a = 7$.
Why did he come to a conclusion where $a$ is $7$, where $x < 1$?
calculus
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
$endgroup$
add a comment |
$begingroup$
I am watching a video trying to understand continuity in calculus. The author first starts with the $ax - 4, x < 1$, and we are trying to find the value of "a".
The author subsequently derives $ax - 4$ as $7x - 4$, where $a = 7$.
Why did he come to a conclusion where $a$ is $7$, where $x < 1$?
calculus
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
$endgroup$
I am watching a video trying to understand continuity in calculus. The author first starts with the $ax - 4, x < 1$, and we are trying to find the value of "a".
The author subsequently derives $ax - 4$ as $7x - 4$, where $a = 7$.
Why did he come to a conclusion where $a$ is $7$, where $x < 1$?
calculus
calculus
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
asked Mar 28 at 15:30
ilovetolearnilovetolearn
63521021
63521021
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
A function is continuous in $x=c$ if its left- and right-hand limits in $c$ are equal.
In the case of $c=1$, this comes down to:
$$lim_x to 1^+colorbluef(x) = lim_x to 1^-colorredf(x) tag$*$$$
Now for the right-hand limit you have $x > 1$ and you need the blue part while for the left-hand limit you have $x < 1$ and you need the red part of the function:
$$f(x)=begincasescolorblue3x^2 & x ge 1\
colorredax-4 & x<1endcases$$
That turns $(*)$ into:
$$lim_x to 1^+left(colorblue3x^2right)= lim_x to 1^-left(colorredax-4right)$$
And then you get:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 28 at 15:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
add a comment |
add a comment |
$begingroup$
A function is continuous in $x=c$ if its left- and right-hand limits in $c$ are equal.
In the case of $c=1$, this comes down to:
$$lim_x to 1^+colorbluef(x) = lim_x to 1^-colorredf(x) tag$*$$$
Now for the right-hand limit you have $x > 1$ and you need the blue part while for the left-hand limit you have $x < 1$ and you need the red part of the function:
$$f(x)=begincasescolorblue3x^2 & x ge 1\
colorredax-4 & x<1endcases$$
That turns $(*)$ into:
$$lim_x to 1^+left(colorblue3x^2right)= lim_x to 1^-left(colorredax-4right)$$
And then you get:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
A function is continuous in $x=c$ if its left- and right-hand limits in $c$ are equal.
In the case of $c=1$, this comes down to:
$$lim_x to 1^+colorbluef(x) = lim_x to 1^-colorredf(x) tag$*$$$
Now for the right-hand limit you have $x > 1$ and you need the blue part while for the left-hand limit you have $x < 1$ and you need the red part of the function:
$$f(x)=begincasescolorblue3x^2 & x ge 1\
colorredax-4 & x<1endcases$$
That turns $(*)$ into:
$$lim_x to 1^+left(colorblue3x^2right)= lim_x to 1^-left(colorredax-4right)$$
And then you get:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
A function is continuous in $x=c$ if its left- and right-hand limits in $c$ are equal.
In the case of $c=1$, this comes down to:
$$lim_x to 1^+colorbluef(x) = lim_x to 1^-colorredf(x) tag$*$$$
Now for the right-hand limit you have $x > 1$ and you need the blue part while for the left-hand limit you have $x < 1$ and you need the red part of the function:
$$f(x)=begincasescolorblue3x^2 & x ge 1\
colorredax-4 & x<1endcases$$
That turns $(*)$ into:
$$lim_x to 1^+left(colorblue3x^2right)= lim_x to 1^-left(colorredax-4right)$$
And then you get:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
$endgroup$
A function is continuous in $x=c$ if its left- and right-hand limits in $c$ are equal.
In the case of $c=1$, this comes down to:
$$lim_x to 1^+colorbluef(x) = lim_x to 1^-colorredf(x) tag$*$$$
Now for the right-hand limit you have $x > 1$ and you need the blue part while for the left-hand limit you have $x < 1$ and you need the red part of the function:
$$f(x)=begincasescolorblue3x^2 & x ge 1\
colorredax-4 & x<1endcases$$
That turns $(*)$ into:
$$lim_x to 1^+left(colorblue3x^2right)= lim_x to 1^-left(colorredax-4right)$$
And then you get:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
answered Mar 28 at 15:37
StackTDStackTD
24.3k2254
24.3k2254
add a comment |
add a comment |
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