Mixing
Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$
$begingroup$
Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$
I know I have to take the derivative of the first equation, but I don’t know what to do after.
calculus
edited Jan 14 at 4:47
coreyman317
772420
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
$endgroup$
add a comment |
$begingroup$
Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$
I know I have to take the derivative of the first equation, but I don’t know what to do after.
calculus
edited Jan 14 at 4:47
coreyman317
772420
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
$endgroup$
$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36
$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37
add a comment |
$begingroup$
Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$
I know I have to take the derivative of the first equation, but I don’t know what to do after.
calculus
edited Jan 14 at 4:47
coreyman317
772420
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
$endgroup$
Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$
I know I have to take the derivative of the first equation, but I don’t know what to do after.
calculus
calculus
edited Jan 14 at 4:47
coreyman317
772420
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
edited Jan 14 at 4:47
coreyman317
772420
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
edited Jan 14 at 4:47
coreyman317
772420
edited Jan 14 at 4:47
coreyman317
772420
edited Jan 14 at 4:47
coreyman317
772420
772420
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
asked Jan 14 at 4:34
Tara AllahdadiTara Allahdadi
1
1
$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36
$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37
add a comment |
$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36
$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37
$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36
$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36
$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37
$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
Then you can use the point-slope formula to find the equation of the tangent lines.
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.
So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
$endgroup$
add a comment |
$begingroup$
The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$
We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.
answered Jan 14 at 4:52
coreyman317coreyman317
772420
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
Then you can use the point-slope formula to find the equation of the tangent lines.
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
Then you can use the point-slope formula to find the equation of the tangent lines.
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
Then you can use the point-slope formula to find the equation of the tangent lines.
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
$endgroup$
For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
Then you can use the point-slope formula to find the equation of the tangent lines.
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
answered Jan 14 at 4:40
Joel PereiraJoel Pereira
75919
75919
add a comment |
add a comment |
$begingroup$
The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.
So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
$endgroup$
add a comment |
$begingroup$
The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.
So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
$endgroup$
add a comment |
$begingroup$
The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.
So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
$endgroup$
The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.
So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
edited Jan 14 at 4:42
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
answered Jan 14 at 4:40
Jimmy SabaterJimmy Sabater
2,582322
2,582322
add a comment |
add a comment |
$begingroup$
The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$
We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.
answered Jan 14 at 4:52
coreyman317coreyman317
772420
$endgroup$
add a comment |
$begingroup$
The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$
We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.
answered Jan 14 at 4:52
coreyman317coreyman317
772420
$endgroup$
add a comment |
$begingroup$
The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$
We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.
answered Jan 14 at 4:52
coreyman317coreyman317
772420
$endgroup$
The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$
We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.
answered Jan 14 at 4:52
coreyman317coreyman317
772420
answered Jan 14 at 4:52
coreyman317coreyman317
772420
answered Jan 14 at 4:52
coreyman317coreyman317
772420
answered Jan 14 at 4:52
coreyman317coreyman317
772420
772420
add a comment |
add a comment |
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$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36
$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37