Question

Prove that there exist only one real nummber X such that e^+x =O Use the lntermediate Vaue Theorem to proue existence, ond the R ollels Theorem to proue the uniqueness

Answer 1

(1) Given Equcernam e = -X >> € + x = 0 Now we take f(x)=e" +20 since et and a is continuous and for each point on IR. 2) fou is continuou and differentiable differentiable on R. since FCO) = € +0=1 >0 f(-)) = ē'-1 =1-i so е = f(o) >0 and f(-1) co and f is continuon and diferenticuble on IR . By intermecliate rulur theoren there is at least one ce lol such that - f(c)=0 Hence fla=etn has at least one real soot S e a -n hus at least one real root is prored. (2) on the contaay__ - f has more than one root - f(a)=0 and f(b)=0 for some attir since f is continuous on [a,b] and differenhilble on la, b) and f(a)=f(b)=0 then By Rolle's theorem there is at least one c E (a,b) such that F'()=0 CDPose ove since flineet it f'(W) = 0_ 5) P -1 which is contraction to eato VREIR .. Hence our supposition is wing --x has only one real wote isponech