How would I go about finding the roots of (e^x-1)-k

Irrerbthist6n

Irrerbthist6n

Answered question

2022-01-14

How would I go about finding the roots of (ex1)karctan(x)=0?

Answer & Explanation

reinosodairyshm

reinosodairyshm

Beginner2022-01-16Added 36 answers

If k>0 
Let 
f(x)=ex1ktan1xf(x)=exk1+x2,f(x)=ex+k(1+x2)2>0 
 f′(x) is an increasing function f(x)>f()>0f(x) is an increasing function. Hence, f(x)=0 will only have one real root at most. As f()=1+kπ2 and f()>0, for one real root by IVT f()<0. Finally this gives 0<k<2π. The one root is x=0. 
For k>2π, both f()> and f()>0 so f(x)=0 will have 0,2,4,... number of real roots. Since x=0 is essentially a root. So there will two real roots if k>2π. Furthermore, since f′′(x)>0, the function f(x) can only have one min, thus eliminates the possibility of having more than two real roots.

RizerMix

RizerMix

Expert2022-01-20Added 656 answers

x=0 is always a solution, of course. If not, type the equation as k=ex1arctan(x) The right side is f (x). The singularity at x=0 can be removed becauselimx0f(x)=1. We also have limxf(x)=2π and limx+f(x)=+ It seems like f(x) is rising. If so, for 2π<k<1 and 1<k< there are two real roots (x=0 and the root of f(x)=k), otherwise there is only x=0.

alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

Let f(x)=ex1karctan(x). We have f(0)=0. If k=0, obviously x=0 is the only solution. If k0, then f(x)=exk1+x2, so that f(x)=0(1+x2)ex=k. Since (1+x2)ex is increasing, goes to 0 at  and goes to  at  we see that f′ has exactly one real zero, say f′(x0)=0. We observe a drop in f on (,x0) and increases on (x0,), in order for f to reach its minimum at x0 Since we are aware of f(0)=0, it must be that x0<0 and x=0 is the only solution.

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