Examen JBB-Borwick_Kimberly

172 days ago by KimberlyB

Exercise 1

a)
var ('u') g(u)=(1-u^2)/(u^2+3) 
       
var ('u') plot((1-(u^2))/((u^2)+3),u,-1,3) 
       
reset() 
       

b)
var('u') Extremes=plot((1-(u^2))/((u^2)+3),u,-1,3).get_minmax_data() Extremes['ymin'] Extremes['ymax'] 
        
0.33332729147989715
0.33332729147989715

c)
diff((1-(u^2))/((u^2)+3),u) 
        
2*(u^2 - 1)*u/(u^2 + 3)^2 - 2*u/(u^2 + 3)
2*(u^2 - 1)*u/(u^2 + 3)^2 - 2*u/(u^2 + 3)

d)
g(u)=(1-u^2)/(u^2+3) dg(u)=(diff(g,u)) dg(1) 
        
-1/2
-1/2
reset() 
       

Exercise 2

a)
var('x') f(x)=2*x^3-4*x^2-6*x+4 
       
var('x') plot(2*x^3-4*x^2-6*x+4,x,-5,5,ymin=-10,ymax=10) 
       

b)
x=var('x') solve(2*x^3-4*x^2-6*x+4,x) 
        
[x == -1/2*(I*sqrt(3) + 1)*(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) -
1/18*(-13*I*sqrt(3) + 13)/(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) + 2/3, x
== -1/2*(-I*sqrt(3) + 1)*(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) -
1/18*(13*I*sqrt(3) + 13)/(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) + 2/3, x
== (1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) + 13/9/(1/9*I*sqrt(3)*sqrt(79)
+ 8/27)^(1/3) + 2/3]
[x == -1/2*(I*sqrt(3) + 1)*(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) - 1/18*(-13*I*sqrt(3) + 13)/(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) + 2/3, x == -1/2*(-I*sqrt(3) + 1)*(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) - 1/18*(13*I*sqrt(3) + 13)/(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) + 2/3, x == (1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) + 13/9/(1/9*I*sqrt(3)*sqrt(79) + 8/27)^(1/3) + 2/3]

c)
f(x)=2*x^3-4*x^2-6*x+4 limit(f(x),x=0) 
        
4
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