how we can prove taht
v²+v-1≤(4v+v*log10v)3 + 4v+v*log10v
also that
4v+v*log10v ≤ (v²+v-1)3 +(v²+v-1)*log10(v²+v-1
2007-01-31 07:14:58 · 4 answers · asked by rania a 1 in Science & Mathematics ➔ Engineering
how we can prove that v²+v-1≤(4v+v*log10v)3 + 4v+v*log10v?
(4v+v*log10v)3 the nomber 3 is cupe
2007-01-31 07:21:36 · update #1
you could probably prove that using math
2007-01-31 07:32:30 · answer #1 · answered by Anonymous · 0⤊ 1⤋
The first thing that comes to mind would be to use mathematical induction. That is:
1) Prove it works for v = 1 (evaluate)
2) Prove that P(k) implies P(k+1); that is, assume it's true for v = k (or any dummy variable), and use that to prove it's true for k+1
Of course, that only proves it for natural number values of v. Not sure how to do otherwise.
2007-01-31 15:45:18 · answer #2 · answered by chiyaniwatori 2 · 0⤊ 0⤋
If you can take the left side of the inequality and divide it by the right side and show that it never exceeds 1, you have your proof. You might have to differentiate and find the max.
2007-01-31 16:11:36 · answer #3 · answered by Gene 7 · 0⤊ 0⤋
ipro
ve you like this:^2+^-1>(4^+^golo1^)3+4^+^*gol01^????
osla taht ^4+^gol#(^*+^1)*gol01(^*+^-1.
2007-01-31 15:37:42 · answer #4 · answered by aura gcranberrymadurogustosocinc 2 · 0⤊ 1⤋