Given $f(x)=-2arctanleft(e^xright)+lnleft(e^x+2right)$ solve $f(x)=0$.
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I want to find the intersection with the $x$ axis of the following function:
$$f(x)=-2arctanleft(e^xright)+lnleft(e^x+2right)$$
I need to find
$$f(x)=0$$
So we have
$$-2arctanleft(e^xright)+lnleft(e^x+2right)=0rightarrow$$
$$lnleft(e^x+2right)=2arctanleft(e^xright)$$
Because of the logarithim, I need to transform the right side into a logarithm as well so that I can eventually "remove" the logarithms and solve the equation, but I don't know how to do that. Any hints?
limits
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up vote
1
down vote
favorite
I want to find the intersection with the $x$ axis of the following function:
$$f(x)=-2arctanleft(e^xright)+lnleft(e^x+2right)$$
I need to find
$$f(x)=0$$
So we have
$$-2arctanleft(e^xright)+lnleft(e^x+2right)=0rightarrow$$
$$lnleft(e^x+2right)=2arctanleft(e^xright)$$
Because of the logarithim, I need to transform the right side into a logarithm as well so that I can eventually "remove" the logarithms and solve the equation, but I don't know how to do that. Any hints?
limits
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to find the intersection with the $x$ axis of the following function:
$$f(x)=-2arctanleft(e^xright)+lnleft(e^x+2right)$$
I need to find
$$f(x)=0$$
So we have
$$-2arctanleft(e^xright)+lnleft(e^x+2right)=0rightarrow$$
$$lnleft(e^x+2right)=2arctanleft(e^xright)$$
Because of the logarithim, I need to transform the right side into a logarithm as well so that I can eventually "remove" the logarithms and solve the equation, but I don't know how to do that. Any hints?
limits
I want to find the intersection with the $x$ axis of the following function:
$$f(x)=-2arctanleft(e^xright)+lnleft(e^x+2right)$$
I need to find
$$f(x)=0$$
So we have
$$-2arctanleft(e^xright)+lnleft(e^x+2right)=0rightarrow$$
$$lnleft(e^x+2right)=2arctanleft(e^xright)$$
Because of the logarithim, I need to transform the right side into a logarithm as well so that I can eventually "remove" the logarithms and solve the equation, but I don't know how to do that. Any hints?
limits
edited Aug 21 at 13:30
greedoid
27.5k93776
27.5k93776
asked Aug 21 at 13:28
Cesare
587210
587210
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3 Answers
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oldest
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Write $t=e^x>0$:
$$f'(x) = -2e^xover 1+e^2x+ e^xover e^x+2=tt^2-2t-3over (t^2+1)(t+2)$$
$$=t(t-3)(t+1)over (t^2+1)(t+2)$$
so for $xleq ln 3$ it is decresing and for $xgeq ln3$ it is increasing, so it has at most two solutions.
Thanks, is differentiating is the only way to go? If so, looks like I have to wait until I know how to properly differentiate.
â Cesare
Aug 21 at 13:38
I don't know. This is only bound for a number of solution to $f(x)=0$. If now you can guess them you are done.
â greedoid
Aug 21 at 13:40
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up vote
0
down vote
Hint: You will Need a numericval method, the two Solutions are given by $$xapprox -0.71157913296093670740$$ or $$xapprox 2.9343041578235969617$$
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up vote
0
down vote
A numerical answer cones from the following steps.
1. Start with $y=e^x=2$.
2. Take ln
3. Divide by 2.
4. Take tangent
5. add 2
6. Go to step 2
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Write $t=e^x>0$:
$$f'(x) = -2e^xover 1+e^2x+ e^xover e^x+2=tt^2-2t-3over (t^2+1)(t+2)$$
$$=t(t-3)(t+1)over (t^2+1)(t+2)$$
so for $xleq ln 3$ it is decresing and for $xgeq ln3$ it is increasing, so it has at most two solutions.
Thanks, is differentiating is the only way to go? If so, looks like I have to wait until I know how to properly differentiate.
â Cesare
Aug 21 at 13:38
I don't know. This is only bound for a number of solution to $f(x)=0$. If now you can guess them you are done.
â greedoid
Aug 21 at 13:40
add a comment |Â
up vote
0
down vote
accepted
Write $t=e^x>0$:
$$f'(x) = -2e^xover 1+e^2x+ e^xover e^x+2=tt^2-2t-3over (t^2+1)(t+2)$$
$$=t(t-3)(t+1)over (t^2+1)(t+2)$$
so for $xleq ln 3$ it is decresing and for $xgeq ln3$ it is increasing, so it has at most two solutions.
Thanks, is differentiating is the only way to go? If so, looks like I have to wait until I know how to properly differentiate.
â Cesare
Aug 21 at 13:38
I don't know. This is only bound for a number of solution to $f(x)=0$. If now you can guess them you are done.
â greedoid
Aug 21 at 13:40
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Write $t=e^x>0$:
$$f'(x) = -2e^xover 1+e^2x+ e^xover e^x+2=tt^2-2t-3over (t^2+1)(t+2)$$
$$=t(t-3)(t+1)over (t^2+1)(t+2)$$
so for $xleq ln 3$ it is decresing and for $xgeq ln3$ it is increasing, so it has at most two solutions.
Write $t=e^x>0$:
$$f'(x) = -2e^xover 1+e^2x+ e^xover e^x+2=tt^2-2t-3over (t^2+1)(t+2)$$
$$=t(t-3)(t+1)over (t^2+1)(t+2)$$
so for $xleq ln 3$ it is decresing and for $xgeq ln3$ it is increasing, so it has at most two solutions.
answered Aug 21 at 13:36
greedoid
27.5k93776
27.5k93776
Thanks, is differentiating is the only way to go? If so, looks like I have to wait until I know how to properly differentiate.
â Cesare
Aug 21 at 13:38
I don't know. This is only bound for a number of solution to $f(x)=0$. If now you can guess them you are done.
â greedoid
Aug 21 at 13:40
add a comment |Â
Thanks, is differentiating is the only way to go? If so, looks like I have to wait until I know how to properly differentiate.
â Cesare
Aug 21 at 13:38
I don't know. This is only bound for a number of solution to $f(x)=0$. If now you can guess them you are done.
â greedoid
Aug 21 at 13:40
Thanks, is differentiating is the only way to go? If so, looks like I have to wait until I know how to properly differentiate.
â Cesare
Aug 21 at 13:38
Thanks, is differentiating is the only way to go? If so, looks like I have to wait until I know how to properly differentiate.
â Cesare
Aug 21 at 13:38
I don't know. This is only bound for a number of solution to $f(x)=0$. If now you can guess them you are done.
â greedoid
Aug 21 at 13:40
I don't know. This is only bound for a number of solution to $f(x)=0$. If now you can guess them you are done.
â greedoid
Aug 21 at 13:40
add a comment |Â
up vote
0
down vote
Hint: You will Need a numericval method, the two Solutions are given by $$xapprox -0.71157913296093670740$$ or $$xapprox 2.9343041578235969617$$
add a comment |Â
up vote
0
down vote
Hint: You will Need a numericval method, the two Solutions are given by $$xapprox -0.71157913296093670740$$ or $$xapprox 2.9343041578235969617$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: You will Need a numericval method, the two Solutions are given by $$xapprox -0.71157913296093670740$$ or $$xapprox 2.9343041578235969617$$
Hint: You will Need a numericval method, the two Solutions are given by $$xapprox -0.71157913296093670740$$ or $$xapprox 2.9343041578235969617$$
answered Aug 21 at 13:41
Dr. Sonnhard Graubner
67.6k32660
67.6k32660
add a comment |Â
add a comment |Â
up vote
0
down vote
A numerical answer cones from the following steps.
1. Start with $y=e^x=2$.
2. Take ln
3. Divide by 2.
4. Take tangent
5. add 2
6. Go to step 2
add a comment |Â
up vote
0
down vote
A numerical answer cones from the following steps.
1. Start with $y=e^x=2$.
2. Take ln
3. Divide by 2.
4. Take tangent
5. add 2
6. Go to step 2
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A numerical answer cones from the following steps.
1. Start with $y=e^x=2$.
2. Take ln
3. Divide by 2.
4. Take tangent
5. add 2
6. Go to step 2
A numerical answer cones from the following steps.
1. Start with $y=e^x=2$.
2. Take ln
3. Divide by 2.
4. Take tangent
5. add 2
6. Go to step 2
answered Aug 21 at 13:51
Empy2
31.9k12059
31.9k12059
add a comment |Â
add a comment |Â
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