#For which values k>0 does the following equation not have a solution?
narrow flame
unborn sinewBOT
limpid oracle
Which number systems is this using?
limpid oracle
ok
well
if ur assuming all reals then its 0
idk if there are any other wierd solutions
but yeah 0 is a pretty on the nose one
cedar tartan
The condition is k>0
limpid oracle
The condition is k>0
i beleive the negative soltuions can be represented using the w lambert function
Oh
wait
its reals
nvm
yeah
above 0 ignore me lol
limpid oracle
The condition is k>0
i just got some free time to think about it
it hass soltuions for k below 0
This is the general soltuion
This will work for ALL a except 0
$x=\frac{5W\left(\frac{k}{15}\right)}{k}$
brave vaultBOT
Sonni
meager path
.solved
unborn sinewBOT
Solved
unborn sinewBOT
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narrow flame
.unsolved
unborn sinewBOT
Unsolved
narrow flame
If k(x) =
Then k’(x) =
We can find posible extremums in in points where k’(x) = 0
For k’(x) to equal 0, 5(1-ln(3x)) must be equal to 0
For 5(1-ln(3x)) to = 0, 1-ln(3x) must = 0
1 - ln(3x) = 0
Solving for x
1 = ln(3x)
e = 3x
e/3 = x
x = e/3 is a root of k’ and a possible extremum for k
A corollary of Bolzanos theorm of intermediate value allows us to predict if the function is going to increase or decrease if we evaluate the sign of k’(x) for a value from the interval ranging from (0, e/3) and a value from the interval (e/3, +inf)
0 is our left-most limit because our domain is limited due to the ln
For (0, e/3) i’ll evaluate k’(1/2)
k’(1/2) ≈ 11.8 (positive therefore k(x) will increase during the interval (0,e/3))
For (e/3, +inf) i’ll evaluate k’(2)
k’(2) ≈ -0.9 (negative therefore k will decrease during the interval (e/3, +inf))
k(x) has an maximum at x = e/3
the range of k(x) is a list of possible values of k for which our original equation has a soultion
Therefore to find the possible values of k>0 for which the equation has a solution we have to limit the range for positive values only
We can do this by finding the roots of k(x)
k(x) = 0 when x = 1/3
Gathering all of the information we can analyze k(x)
k(x) increases from x = 0 to x = e/3 at which point it has a maximum and then decreases towards +inf
Once again Bolzanos theorm lets us predict that values of k before x = 1/3 (root of k) are negative and the ones following are positive
Therefore the range of possible values for k>0 for which our original equation has a solution = (0, e/3]
We can verify by using geogebra to graph
k(x) graph
black x = e/3 (extremum, root of k’)
Root of k(x)
cinder ether
It seems like solutions exist for k > e/3, no?
narrow flame
k’(x) graph