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Finding the slope of the line normal to the graph at a given point
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My Problem: What is the slope of the line normal to the graph of $f(x) = e^x-x^e-e$ at the point where the graph crosses the $x$-axis?
- a. $-0.288$
- b. $-0.110$
- c. $3.471$
- d. $9.106$
I found that $f(x)$ crosses the $x$ axis at $3.471$, but is that the answer? Do I have to plug it back into the equation? Thank you!
calculus derivatives slope
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
asked Jan 21 at 16:00
MichelleMichelle
1
$endgroup$
add a comment |
$begingroup$
My Problem: What is the slope of the line normal to the graph of $f(x) = e^x-x^e-e$ at the point where the graph crosses the $x$-axis?
- a. $-0.288$
- b. $-0.110$
- c. $3.471$
- d. $9.106$
I found that $f(x)$ crosses the $x$ axis at $3.471$, but is that the answer? Do I have to plug it back into the equation? Thank you!
calculus derivatives slope
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
asked Jan 21 at 16:00
MichelleMichelle
1
$endgroup$
$begingroup$
This is definitively not the searched answer, you must compute the slope at this point and then the slope of the normal line
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 17:56
add a comment |
$begingroup$
My Problem: What is the slope of the line normal to the graph of $f(x) = e^x-x^e-e$ at the point where the graph crosses the $x$-axis?
- a. $-0.288$
- b. $-0.110$
- c. $3.471$
- d. $9.106$
I found that $f(x)$ crosses the $x$ axis at $3.471$, but is that the answer? Do I have to plug it back into the equation? Thank you!
calculus derivatives slope
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
asked Jan 21 at 16:00
MichelleMichelle
1
$endgroup$
My Problem: What is the slope of the line normal to the graph of $f(x) = e^x-x^e-e$ at the point where the graph crosses the $x$-axis?
- a. $-0.288$
- b. $-0.110$
- c. $3.471$
- d. $9.106$
I found that $f(x)$ crosses the $x$ axis at $3.471$, but is that the answer? Do I have to plug it back into the equation? Thank you!
calculus derivatives slope
calculus derivatives slope
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
asked Jan 21 at 16:00
MichelleMichelle
1
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
asked Jan 21 at 16:00
MichelleMichelle
1
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
edited Jan 21 at 16:11
Eevee Trainer
7,52721338
7,52721338
asked Jan 21 at 16:00
MichelleMichelle
1
asked Jan 21 at 16:00
MichelleMichelle
1
asked Jan 21 at 16:00
MichelleMichelle
1
1
$begingroup$
This is definitively not the searched answer, you must compute the slope at this point and then the slope of the normal line
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 17:56
add a comment |
$begingroup$
This is definitively not the searched answer, you must compute the slope at this point and then the slope of the normal line
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 17:56
$begingroup$
This is definitively not the searched answer, you must compute the slope at this point and then the slope of the normal line
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 17:56
$begingroup$
This is definitively not the searched answer, you must compute the slope at this point and then the slope of the normal line
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 17:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Recall: the equation of the line at $(x^*,y^*)$ for a function $f$ can be given by the point-slope form:
$$y - y^* = f'(x^*) cdot (x - x^*)$$
(analogous to the $y=mx+b$ form for linear equations). The point where $f$ crosses the $x$ axis is a point where $y^* = 0$, and you found $x^* approx 3.471$. All that remains for this part is to find $f'$ and then plug in everything as necessary above.
Shift it into the $y = mx+b$ form, then, where $m$ will be the slope of the tangent line at that point.
From there, recall that the "normal" line is just a line perpendicular to the given line. Suppose we have two lines,
$$y_1 = m_1 x + b_1 ;;;;; y_2 = m_2 x + b_2$$
If $y_2$ is perpendicular to $y_1$, then
$$m_2 = - frac{1}{m_1}$$
i.e. the slopes of perpendicular lines are negative reciprocals of each other.
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
$endgroup$
add a comment |
$begingroup$
Hint: We get by the sum and the power rule $$f'(x)={{rm e}^{x}}-{frac {{x}^{{rm e}}{rm e}}{x}}$$ Can you proceed?
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Recall: the equation of the line at $(x^*,y^*)$ for a function $f$ can be given by the point-slope form:
$$y - y^* = f'(x^*) cdot (x - x^*)$$
(analogous to the $y=mx+b$ form for linear equations). The point where $f$ crosses the $x$ axis is a point where $y^* = 0$, and you found $x^* approx 3.471$. All that remains for this part is to find $f'$ and then plug in everything as necessary above.
Shift it into the $y = mx+b$ form, then, where $m$ will be the slope of the tangent line at that point.
From there, recall that the "normal" line is just a line perpendicular to the given line. Suppose we have two lines,
$$y_1 = m_1 x + b_1 ;;;;; y_2 = m_2 x + b_2$$
If $y_2$ is perpendicular to $y_1$, then
$$m_2 = - frac{1}{m_1}$$
i.e. the slopes of perpendicular lines are negative reciprocals of each other.
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
$endgroup$
add a comment |
$begingroup$
Recall: the equation of the line at $(x^*,y^*)$ for a function $f$ can be given by the point-slope form:
$$y - y^* = f'(x^*) cdot (x - x^*)$$
(analogous to the $y=mx+b$ form for linear equations). The point where $f$ crosses the $x$ axis is a point where $y^* = 0$, and you found $x^* approx 3.471$. All that remains for this part is to find $f'$ and then plug in everything as necessary above.
Shift it into the $y = mx+b$ form, then, where $m$ will be the slope of the tangent line at that point.
From there, recall that the "normal" line is just a line perpendicular to the given line. Suppose we have two lines,
$$y_1 = m_1 x + b_1 ;;;;; y_2 = m_2 x + b_2$$
If $y_2$ is perpendicular to $y_1$, then
$$m_2 = - frac{1}{m_1}$$
i.e. the slopes of perpendicular lines are negative reciprocals of each other.
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
$endgroup$
add a comment |
$begingroup$
Recall: the equation of the line at $(x^*,y^*)$ for a function $f$ can be given by the point-slope form:
$$y - y^* = f'(x^*) cdot (x - x^*)$$
(analogous to the $y=mx+b$ form for linear equations). The point where $f$ crosses the $x$ axis is a point where $y^* = 0$, and you found $x^* approx 3.471$. All that remains for this part is to find $f'$ and then plug in everything as necessary above.
Shift it into the $y = mx+b$ form, then, where $m$ will be the slope of the tangent line at that point.
From there, recall that the "normal" line is just a line perpendicular to the given line. Suppose we have two lines,
$$y_1 = m_1 x + b_1 ;;;;; y_2 = m_2 x + b_2$$
If $y_2$ is perpendicular to $y_1$, then
$$m_2 = - frac{1}{m_1}$$
i.e. the slopes of perpendicular lines are negative reciprocals of each other.
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
$endgroup$
Recall: the equation of the line at $(x^*,y^*)$ for a function $f$ can be given by the point-slope form:
$$y - y^* = f'(x^*) cdot (x - x^*)$$
(analogous to the $y=mx+b$ form for linear equations). The point where $f$ crosses the $x$ axis is a point where $y^* = 0$, and you found $x^* approx 3.471$. All that remains for this part is to find $f'$ and then plug in everything as necessary above.
Shift it into the $y = mx+b$ form, then, where $m$ will be the slope of the tangent line at that point.
From there, recall that the "normal" line is just a line perpendicular to the given line. Suppose we have two lines,
$$y_1 = m_1 x + b_1 ;;;;; y_2 = m_2 x + b_2$$
If $y_2$ is perpendicular to $y_1$, then
$$m_2 = - frac{1}{m_1}$$
i.e. the slopes of perpendicular lines are negative reciprocals of each other.
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
answered Jan 21 at 16:09
Eevee TrainerEevee Trainer
7,52721338
7,52721338
add a comment |
add a comment |
$begingroup$
Hint: We get by the sum and the power rule $$f'(x)={{rm e}^{x}}-{frac {{x}^{{rm e}}{rm e}}{x}}$$ Can you proceed?
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
add a comment |
$begingroup$
Hint: We get by the sum and the power rule $$f'(x)={{rm e}^{x}}-{frac {{x}^{{rm e}}{rm e}}{x}}$$ Can you proceed?
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
add a comment |
$begingroup$
Hint: We get by the sum and the power rule $$f'(x)={{rm e}^{x}}-{frac {{x}^{{rm e}}{rm e}}{x}}$$ Can you proceed?
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
Hint: We get by the sum and the power rule $$f'(x)={{rm e}^{x}}-{frac {{x}^{{rm e}}{rm e}}{x}}$$ Can you proceed?
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
answered Jan 21 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
77k42866
add a comment |
add a comment |
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$begingroup$
This is definitively not the searched answer, you must compute the slope at this point and then the slope of the normal line
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 17:56