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Calculus: Finding an unknown coefficient in continuous equations

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0

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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

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asked Mar 28 at 15:30

ilovetolearnilovetolearn

63521021

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add a comment | 

0

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked Mar 28 at 15:30

ilovetolearnilovetolearn

63521021

$endgroup$

share|cite|improve this question

asked Mar 28 at 15:30

ilovetolearnilovetolearn

63521021

share|cite|improve this question

asked Mar 28 at 15:30

ilovetolearnilovetolearn

63521021

asked Mar 28 at 15:30

ilovetolearnilovetolearn

63521021

asked Mar 28 at 15:30

ilovetolearnilovetolearn

63521021

2 Answers
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3

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It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.

share|cite|improve this answer

answered Mar 28 at 15:34

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

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add a comment | 

1

$begingroup$

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:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$

share|cite|improve this answer

answered Mar 28 at 15:37

StackTDStackTD

24.3k2254

$endgroup$

add a comment | 

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3

$begingroup$

It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.

share|cite|improve this answer

answered Mar 28 at 15:34

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

3

$begingroup$

It is right, since $$lim_xto 1^-ax-4=a-4$$ and $$f(1)=1$$ so we get
$$a-4=3$$ as you stated.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 28 at 15:34

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

share|cite|improve this answer

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

1

$begingroup$

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:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$

share|cite|improve this answer

answered Mar 28 at 15:37

StackTDStackTD

24.3k2254

$endgroup$

add a comment | 

1

$begingroup$

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:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$

share|cite|improve this answer

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:
$$3cdot 1^2 = a cdot 1 -4 iff 3=a-4 iff a = 7$$

share|cite|improve this answer

answered Mar 28 at 15:37

StackTDStackTD

24.3k2254

$endgroup$

share|cite|improve this answer

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