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How to find parallel tangent lines

Finding equations of tangent lines that are parallelTangent lines to curve parallel to $x-4y = 2$Finding a tangent to an ellipse parallel to a given lineUse implicit differentiation to find tangent lines parallel to a given lineBasic Derivatives-finding tangent linesHow many lines tangent to the graph of $f$ are parallel to the lineEquation of two parallel tangent lines and coordinatesFind Equations of tangent linesFinding where two graphs have perpendicular tangent linesFind specific tangent lines of $y=2x^3-3x^2-12x+20$

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5

$begingroup$

I was given this problem:

Let $f$ be the function defined by $f(x)=e^{3x}$, and let $g$ be the function defined by $g(x)=x^3$. At what value of $x$ do the graphs of $f$ and $g$ have parallel tangent lines?

To solve this problem, I set the derivatives of $f$ and $g$ equal to each other: $3x^2=3e^{3x}$. Now, I just need to solve for $x$, but for some reason I am completely at a loss of how to do that. I divided the problem by $3$, which got me $x^2=e^{3x}$. But, now what?

I also plugged it into my calculators solver, which got me the correct answer: $-.484$. But, I don't know how to find that answer without solver.

calculus derivatives

share|cite|improve this question

edited 7 hours ago

Adrian Keister

6,37277 gold badges2222 silver badges3333 bronze badges

asked 8 hours ago

burtburt

1391010 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

I was given this problem:

Let $f$ be the function defined by $f(x)=e^{3x}$, and let $g$ be the function defined by $g(x)=x^3$. At what value of $x$ do the graphs of $f$ and $g$ have parallel tangent lines?

To solve this problem, I set the derivatives of $f$ and $g$ equal to each other: $3x^2=3e^{3x}$. Now, I just need to solve for $x$, but for some reason I am completely at a loss of how to do that. I divided the problem by $3$, which got me $x^2=e^{3x}$. But, now what?

I also plugged it into my calculators solver, which got me the correct answer: $-.484$. But, I don't know how to find that answer without solver.

calculus derivatives

share|cite|improve this question

edited 7 hours ago

Adrian Keister

6,37277 gold badges2222 silver badges3333 bronze badges

asked 8 hours ago

burtburt

1391010 bronze badges

$endgroup$

add a comment | 

$begingroup$

I was given this problem:

Let $f$ be the function defined by $f(x)=e^{3x}$, and let $g$ be the function defined by $g(x)=x^3$. At what value of $x$ do the graphs of $f$ and $g$ have parallel tangent lines?

To solve this problem, I set the derivatives of $f$ and $g$ equal to each other: $3x^2=3e^{3x}$. Now, I just need to solve for $x$, but for some reason I am completely at a loss of how to do that. I divided the problem by $3$, which got me $x^2=e^{3x}$. But, now what?

I also plugged it into my calculators solver, which got me the correct answer: $-.484$. But, I don't know how to find that answer without solver.

calculus derivatives

share|cite|improve this question

edited 7 hours ago

Adrian Keister

6,37277 gold badges2222 silver badges3333 bronze badges

asked 8 hours ago

burtburt

1391010 bronze badges

$endgroup$

share|cite|improve this question

edited 7 hours ago

Adrian Keister

6,37277 gold badges2222 silver badges3333 bronze badges

asked 8 hours ago

burtburt

1391010 bronze badges

share|cite|improve this question

edited 7 hours ago

Adrian Keister

6,37277 gold badges2222 silver badges3333 bronze badges

asked 8 hours ago

burtburt

1391010 bronze badges

edited 7 hours ago

Adrian Keister

6,37277 gold badges2222 silver badges3333 bronze badges

edited 7 hours ago

Adrian Keister

6,37277 gold badges2222 silver badges3333 bronze badges

asked 8 hours ago

burtburt

1391010 bronze badges

asked 8 hours ago

burtburt

1391010 bronze badges

2 Answers
2

active

oldest

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3

$begingroup$

This problem can be solved using the Lambert-W function as follows.
$$x^2=e^{3x}$$
$$sqrt{x^2}=sqrt{e^{3x}}$$
$$x=pm e^{3x/2}$$
$$-frac32 x=mpfrac32 e^{3x/2}$$
$$-frac32 xe^{-3x/2}=mpfrac32$$
$$-frac32 x=W_kleft(mpfrac32right)$$
$$x=-frac23W_kleft(pmfrac32right)$$
for any branch of the function $kinmathbb{Z}$. The only real solution would be when $k=0$ and $+$ is taken for the $pm$ giving
$$x=-frac23W_0left(frac32right)approx-0.4839075718dots$$
which is the value provided by your calculator.

share|cite|improve this answer

answered 7 hours ago

Peter ForemanPeter Foreman

12.8k11 gold badge55 silver badges2828 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Since the other answers provide the exact solution using the Lambert $W$ function, let me show you how you can prove that the equation $x^2 = e^{3x}$ has one and only one solution using elementary methods.

Let $F(x) = x^2$ and $G(x) = e^{3x}$.

If you plot $F$ and $G$ on the same plane, you can see that there is exactly one point $x$ such that $F(x) = G(x)$. Here $y = F(x)$ is in red and $y = G(x)$ is in blue:

Now, if you want to prove that there is a unique solution to the equation, you can reason as follows.

Consider first the interval $(-infty, 0]$.

1. Since $F$ is decreasing and $G$ is increasing, there is at most one $x in (-infty, 0]$ such that $F(x) = G(x)$.


2. Since $F(-1) > G(-1)$ and $F(0) < G(0)$, there is at least one $x in (-1, 0)$ such that $F(x) = G(x)$.

Combining the two results, we know that there is exactly one $x in (-infty, 0]$ such that $F(x) = G(x)$. Moreover, $x in (-1, 0)$.

For the interval $(0, infty)$ the situation is a bit trickier, because both functions are increasing. But if you know that $x < e^x$ for any $x > 0$, you can prove that $F(x) < G(x)$ as follows:
$$x^2 < (e^x)^2 = e^{2x} < e^{3x}$$
Therefore there are no solutions in $(0, infty)$.

share|cite|improve this answer

answered 7 hours ago

Luca BressanLuca Bressan

4,29022 gold badges1010 silver badges3838 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So would I be able to find the answer on a calculator by graphing the two equations and finding the intersection?
  $endgroup$
  – burt
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If you don't want (or aren't allowed) to use a graphing calculator, you can still plot the two graphs by hand. For a better approximation of $x$, you just need to find a sufficiently small interval $(a, b)$ such that $F(a) > G(a)$ and $F(b) < G(b)$, e.g. $(-0.49, -0.48)$.
  $endgroup$
  – Luca Bressan
  7 hours ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

This problem can be solved using the Lambert-W function as follows.
$$x^2=e^{3x}$$
$$sqrt{x^2}=sqrt{e^{3x}}$$
$$x=pm e^{3x/2}$$
$$-frac32 x=mpfrac32 e^{3x/2}$$
$$-frac32 xe^{-3x/2}=mpfrac32$$
$$-frac32 x=W_kleft(mpfrac32right)$$
$$x=-frac23W_kleft(pmfrac32right)$$
for any branch of the function $kinmathbb{Z}$. The only real solution would be when $k=0$ and $+$ is taken for the $pm$ giving
$$x=-frac23W_0left(frac32right)approx-0.4839075718dots$$
which is the value provided by your calculator.

share|cite|improve this answer

answered 7 hours ago

Peter ForemanPeter Foreman

12.8k11 gold badge55 silver badges2828 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

This problem can be solved using the Lambert-W function as follows.
$$x^2=e^{3x}$$
$$sqrt{x^2}=sqrt{e^{3x}}$$
$$x=pm e^{3x/2}$$
$$-frac32 x=mpfrac32 e^{3x/2}$$
$$-frac32 xe^{-3x/2}=mpfrac32$$
$$-frac32 x=W_kleft(mpfrac32right)$$
$$x=-frac23W_kleft(pmfrac32right)$$
for any branch of the function $kinmathbb{Z}$. The only real solution would be when $k=0$ and $+$ is taken for the $pm$ giving
$$x=-frac23W_0left(frac32right)approx-0.4839075718dots$$
which is the value provided by your calculator.

share|cite|improve this answer

answered 7 hours ago

Peter ForemanPeter Foreman

12.8k11 gold badge55 silver badges2828 bronze badges

$endgroup$

add a comment | 

$begingroup$

This problem can be solved using the Lambert-W function as follows.
$$x^2=e^{3x}$$
$$sqrt{x^2}=sqrt{e^{3x}}$$
$$x=pm e^{3x/2}$$
$$-frac32 x=mpfrac32 e^{3x/2}$$
$$-frac32 xe^{-3x/2}=mpfrac32$$
$$-frac32 x=W_kleft(mpfrac32right)$$
$$x=-frac23W_kleft(pmfrac32right)$$
for any branch of the function $kinmathbb{Z}$. The only real solution would be when $k=0$ and $+$ is taken for the $pm$ giving
$$x=-frac23W_0left(frac32right)approx-0.4839075718dots$$
which is the value provided by your calculator.

share|cite|improve this answer

answered 7 hours ago

Peter ForemanPeter Foreman

12.8k11 gold badge55 silver badges2828 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Peter ForemanPeter Foreman

12.8k11 gold badge55 silver badges2828 bronze badges

answered 7 hours ago

Peter ForemanPeter Foreman

12.8k11 gold badge55 silver badges2828 bronze badges

answered 7 hours ago

Peter ForemanPeter Foreman

12.8k11 gold badge55 silver badges2828 bronze badges

3

$begingroup$

Since the other answers provide the exact solution using the Lambert $W$ function, let me show you how you can prove that the equation $x^2 = e^{3x}$ has one and only one solution using elementary methods.

Let $F(x) = x^2$ and $G(x) = e^{3x}$.

If you plot $F$ and $G$ on the same plane, you can see that there is exactly one point $x$ such that $F(x) = G(x)$. Here $y = F(x)$ is in red and $y = G(x)$ is in blue:

Now, if you want to prove that there is a unique solution to the equation, you can reason as follows.

Consider first the interval $(-infty, 0]$.

1. Since $F$ is decreasing and $G$ is increasing, there is at most one $x in (-infty, 0]$ such that $F(x) = G(x)$.


2. Since $F(-1) > G(-1)$ and $F(0) < G(0)$, there is at least one $x in (-1, 0)$ such that $F(x) = G(x)$.

Combining the two results, we know that there is exactly one $x in (-infty, 0]$ such that $F(x) = G(x)$. Moreover, $x in (-1, 0)$.

For the interval $(0, infty)$ the situation is a bit trickier, because both functions are increasing. But if you know that $x < e^x$ for any $x > 0$, you can prove that $F(x) < G(x)$ as follows:
$$x^2 < (e^x)^2 = e^{2x} < e^{3x}$$
Therefore there are no solutions in $(0, infty)$.

share|cite|improve this answer

answered 7 hours ago

Luca BressanLuca Bressan

4,29022 gold badges1010 silver badges3838 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So would I be able to find the answer on a calculator by graphing the two equations and finding the intersection?
  $endgroup$
  – burt
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If you don't want (or aren't allowed) to use a graphing calculator, you can still plot the two graphs by hand. For a better approximation of $x$, you just need to find a sufficiently small interval $(a, b)$ such that $F(a) > G(a)$ and $F(b) < G(b)$, e.g. $(-0.49, -0.48)$.
  $endgroup$
  – Luca Bressan
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Since the other answers provide the exact solution using the Lambert $W$ function, let me show you how you can prove that the equation $x^2 = e^{3x}$ has one and only one solution using elementary methods.

Let $F(x) = x^2$ and $G(x) = e^{3x}$.

If you plot $F$ and $G$ on the same plane, you can see that there is exactly one point $x$ such that $F(x) = G(x)$. Here $y = F(x)$ is in red and $y = G(x)$ is in blue:

Now, if you want to prove that there is a unique solution to the equation, you can reason as follows.

Consider first the interval $(-infty, 0]$.

1. Since $F$ is decreasing and $G$ is increasing, there is at most one $x in (-infty, 0]$ such that $F(x) = G(x)$.


2. Since $F(-1) > G(-1)$ and $F(0) < G(0)$, there is at least one $x in (-1, 0)$ such that $F(x) = G(x)$.

Combining the two results, we know that there is exactly one $x in (-infty, 0]$ such that $F(x) = G(x)$. Moreover, $x in (-1, 0)$.

For the interval $(0, infty)$ the situation is a bit trickier, because both functions are increasing. But if you know that $x < e^x$ for any $x > 0$, you can prove that $F(x) < G(x)$ as follows:
$$x^2 < (e^x)^2 = e^{2x} < e^{3x}$$
Therefore there are no solutions in $(0, infty)$.

share|cite|improve this answer

answered 7 hours ago

Luca BressanLuca Bressan

4,29022 gold badges1010 silver badges3838 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So would I be able to find the answer on a calculator by graphing the two equations and finding the intersection?
  $endgroup$
  – burt
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If you don't want (or aren't allowed) to use a graphing calculator, you can still plot the two graphs by hand. For a better approximation of $x$, you just need to find a sufficiently small interval $(a, b)$ such that $F(a) > G(a)$ and $F(b) < G(b)$, e.g. $(-0.49, -0.48)$.
  $endgroup$
  – Luca Bressan
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Since the other answers provide the exact solution using the Lambert $W$ function, let me show you how you can prove that the equation $x^2 = e^{3x}$ has one and only one solution using elementary methods.

Let $F(x) = x^2$ and $G(x) = e^{3x}$.

If you plot $F$ and $G$ on the same plane, you can see that there is exactly one point $x$ such that $F(x) = G(x)$. Here $y = F(x)$ is in red and $y = G(x)$ is in blue:

Now, if you want to prove that there is a unique solution to the equation, you can reason as follows.

Consider first the interval $(-infty, 0]$.

1. Since $F$ is decreasing and $G$ is increasing, there is at most one $x in (-infty, 0]$ such that $F(x) = G(x)$.


2. Since $F(-1) > G(-1)$ and $F(0) < G(0)$, there is at least one $x in (-1, 0)$ such that $F(x) = G(x)$.

Combining the two results, we know that there is exactly one $x in (-infty, 0]$ such that $F(x) = G(x)$. Moreover, $x in (-1, 0)$.

For the interval $(0, infty)$ the situation is a bit trickier, because both functions are increasing. But if you know that $x < e^x$ for any $x > 0$, you can prove that $F(x) < G(x)$ as follows:
$$x^2 < (e^x)^2 = e^{2x} < e^{3x}$$
Therefore there are no solutions in $(0, infty)$.

share|cite|improve this answer

answered 7 hours ago

Luca BressanLuca Bressan

4,29022 gold badges1010 silver badges3838 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Luca BressanLuca Bressan

4,29022 gold badges1010 silver badges3838 bronze badges

answered 7 hours ago

Luca BressanLuca Bressan

4,29022 gold badges1010 silver badges3838 bronze badges

answered 7 hours ago

Luca BressanLuca Bressan

4,29022 gold badges1010 silver badges3838 bronze badges