ojzha的博客

T 4 解： (1) x=5−x32x = \frac{5-x^3}{2}x=25−x3​ ϕ(x)=5−x32\phi(x) = \frac{5-x^3}{2}ϕ(x)=25−x3​ ,当x ∈[1,2]\in [1,2]∈[1,2]时有: ϕ′(x)=−3x22<0 \phi^{'}(x) = -\frac{3x^{2}}{2} < 0 ϕ′(x)=−23x2​<0 ∣ϕ′(x)∣=∣3x22∣≥3∗132=1.5>1 \lvert \phi^{'}(x) \rv