제2장함수와그래프

2018 days 전, suziejjang 작성

f(x)=x+3 plot(f(x), (x, -5, 5), ymax=10) 
       
f(x)=(x+2)^2 plot(f(x), (x, -5, 5), ymax=30) 
       
f(x)=abs(x)-4 plot(f, (x, -6, 6)) 
       
f(x)=(x+1)^3 plot(f(x), (x, -3, 3), ymax=30) 
       
f(x)=sqrt(x-5) plot(f(x), (x, -10, 10), ymax=4) 
       
verbose 0 (2395: plot.py, generate_plot_points) WARNING: When plotting,
failed to evaluate function at 149 points.
verbose 0 (2395: plot.py, generate_plot_points) Last error message: ''
verbose 0 (2395: plot.py, generate_plot_points) WARNING: When plotting, failed to evaluate function at 149 points.
verbose 0 (2395: plot.py, generate_plot_points) Last error message: ''
f(x)=(x-1)^2-3 plot(f(x), (x, -8, 8), ymax=15) 
       

2.반사의 성질을 적용하여 다음 그래프를 그리고 감소 및 증가 구간을 정하여라(문제1참고)

f(x)= x+3 g(x)= -x-3 plot(f(x), (x, -8, 8))+plot(g(x), (x, -8, 8), color='red') 
       
f(x)= (x+2)^2 g(x)= -(x+2)^2 plot(f(x), (x, -20, 20))+plot(g(x), (x, -20,20), color='red') 
       
f(x)= abs(x)-4 g(x)= -abs(x)-4 plot(f(x), (x, -15, 15))+plot(g(x), (x, -15,15), color='red') 
       
f(x)=(x+1)^3 g(x)=-(x+1)^3 plot(f(x), (x, -5, 3), ymax=30 , color='blue')+ plot(g(x), (x, -5,3),ymax=30, color='red') 
       
f(x)=sqrt(x-5) g(x)=-sqrt(x-5) plot(f(x), (x, 5, 30), ymax=10, color='blue')+ plot(g(x), (x, 5,30),ymax=10, color='red') 
       
f(x)=(x-1)^2-3 g(x)=-(x-1)^2-3 plot(f(x), (x, -10, 10), ymax=15, color='blue')+ plot(g(x), (x, -10, 10), ymax=15,color='red')