Thus, we obtain 0 < y ( t 1 + ) < y ( t 2 ) = y ( t 2 − ), y ′ ( t 2 ) = y ′ ( t 2 − ) ≥ 0. We use the preceding procedure and deduce by induction that y ( t k ) > 0, y ′ ( t k ) ≥ 0, k = 1, 2, …, p + 1, which contradicts that y ( 1 ) cannot be the maximum point.
By virtue of Step 5, it is clear that one may assume the sequence ({ x_{varepsilon}}) in Step 1 to be the maximum points of (u_{varepsilon}).
A cut off of 7 points was arbitrarily selected as the mid-point between 0 and 14 which was the maximum point achieved by households.
Moreover, from (F3) one easily deduces that the critical point of g ( u, v ) is unique in ( 0, ∞ ), and then it is the maximum point.
we get, for and, for, thus the operator is monotone increasing on and monotone decreasing on and also is the maximum point of the operator.
They showed that u ε attains its maximum value at a point P ε ∈ ∂ Ω, and the subsequences of P ε converge to P, which is the maximum point of mean curvature on ∂ Ω.
Thus, we conclude that x 0 is the maximum point in the interval ( 0, 1 ) and consequently, we have m p, q = inf x ∈ [ 0, 1 ] l p, q ( x ) = 0 (22).
(5) If (frac{lambda a}{mutheta^{2}}< b0) such that frac{p'(thetasqrt[3]{x^{p'(sqrt[3]{x^=frac{lambda a}{mu theta b}, where (x^) is the maximum point of (g(x)).
If (frac{lambda a}{mutheta^{2}}< b0) such that frac{p'(thetasqrt[3]{x^{p'(sqrt[3]{x^=frac{lambda a}{mu theta b}, where (x^) is the maximum point of (g(x)).
