Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 128896 by ZiYangLee last updated on 11/Jan/21

Prove that even obtaining the zero(s), the following equation has only one zero. f(t)=(1+(√2)t)(1−t^2 )+t^2 (t+(√2))

Provethatevenobtainingthezero(s),thefollowingequationhasonlyonezero.f(t)=(1+2t)(1−t2)+t2(t+2)

Answered by Olaf last updated on 11/Jan/21

f(t) = (1+(√2)t)(1−t^2 )+t^2 (t+(√2)) f′(t) = (√2)(1−t^2 )−2t(1+(√2)t)+2t(t+(√2))+t^2 f′(t) = 3((√2)−1)t^2 +2((√2)−1)t+(√2) Δ = 2^2 ((√2)−1)^2 −4×3×((√2)−1)(√2) = 4(√2)−12 < 0 The sign of f′ is the sign of 3((√2)−1) > 0 ⇒ f is a strictly increasing function and lim_(x→−∞) f(x) = lim_(x→−∞) t^3 = −∞ and lim_(x→+∞) f(x) = lim_(x→+∞) t^3 = +∞ ⇒ only one real root.

f(t)=(1+2t)(1−t2)+t2(t+2)f′(t)=2(1−t2)−2t(1+2t)+2t(t+2)+t2f′(t)=3(2−1)t2+2(2−1)t+2Δ=22(2−1)2−4×3×(2−1)2=42−12<0Thesignoff′isthesignof3(2−1)>0⇒fisastrictlyincreasingfunctionandlimx→−∞f(x)=limx→−∞t3=−∞andlimx→+∞f(x)=limx→+∞t3=+∞⇒onlyonerealroot.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com