Help pls, the text goes 'the equation has at least two solutions'

Draw the graphs the solutions become obvious

Dude, literally just work through basic German grammar book. Language is more of a matter of repetition instead of logical thinking.

I'm trying, can't make it work

Kid, you need to experience failure. Luckily you're in German system, which is very forgiving to failures such as yourself.

I'm austrian friendo, I appreciate your advice but I'd appreciate more if you could throw me a bone and help a bit

Shouldn't you do something like take the derivative, show it has 2 extremas, one negative one positive
Which will show that the function goes through the x axis 3 times, therefore 3 solutions
Or something like that kek

Plot them. Literally just plot them.

that one seems pretty easy actually, not going to help you cheat, but it doesn't converge right? or is it more complicated than I think it is?

x=0 is a double solution

/thread

Also, why do you vote for Merkel?

What about the +1 at the end, shit cant be 0

9 minutes, can anyone help me do these

I'm austrian my dude, I can't vote for merkel

>I can't vote for merkel
>implying that he would if he could

The solutions are -0.236181 and 0.696521

You just need to plot the graph of the functions

[math] f(x) = 7^{|x|} [/math]
[math] g(x) = (x+1)^{2} + 1 [/math]

Use the integral test for the convergence of the series. That is, find out for which values of the greek letters is this integral

[math] \int_2^\infty \frac{1}{x^\beta (\ln x)^{\alpha}} \mathrm{d}x [/math]

To compute the integral use the substitution
[math] t = \ln {x} [/math]
[math] " \mathrm{d}t = \frac{1}{x} \mathrm{d}x "[/math]

Consider
[math]f(x) = 7^{|x|} - (x+1)^2 - 1 [/math]
and notice that
[math]f(-1) = 6 [/math]
[math]f(0) = -1 [/math]
[math]f(1) = 2 [/math]
So by the immediate value theorem there is at least one solution in the interval [math]]-1,0[ [/math]and at least one solution in the interval [math]]0,1[ [/math].

For [math] \beta >1 [/math]
[eqn] \sum_{n=2}^\infty \frac{1}{n^\beta (\ln n)^\alpha} < \sum_{n=2}^\infty \frac{1}{n^\beta} < \infty [/eqn]
by the integral test.

just realized I read it as less than one, rather than greater than one. Still stupid easy though, OP has no excuse

Both works. It really depends if you can do that but in this case you can.

Apropos, Mathematik I oder Oberstufen Mathe?
Lösungen hast du ja einige hier bekommen :^)

Notfalls einfach in Wolfram eintippen und so Punkte abkassieren

Und wie kriegst du das hin in der Klausur zu sitzen und gleichzeitig nach Hilfe zu frgaen und zu tippen?

Thanks my dude, you were really helpful, I wish someone helps you too when you need it like you did for me

Actually the answer for the second problem was not complete.
By Cauchy condensation test
[eqn] \sum_{n=2}^\infty \frac{1}{n^\beta (\ln n)^\alpha} [/eqn]
converges iff
[eqn] \sum_{n=2}^\infty 2^n \frac{1}{\left( 2^n \right)^\beta (\ln { \left( 2^n \right))^\alpha}} = \sum_{n=2}^\infty 2^{n (1 - \beta)} n^{-\alpha} \left( \ln 2 \right)^{-\alpha}[/eqn]
converges.

So it converges if either [math] \beta > 1 [/math] or
[math] \beta =1 [/math] and [math] \alpha > 1 [/math] and diverges in all other cases.

I love Veeky Forums. They help OP now so he stays dumb and doesn't learn his lesson.

>helping a brainlet to pass his baby exam

Why you doing this
He doesn't deserve to pass