Question
show that the equation xcosx-2x2+3x-1=0 has at least one solution in [0.2,0.3]
Answers
Jennifer
xcosx-2x2+3x-1=0
Evaluate the function at 0.2 and 0.3:
0.2*cos0.2-2*0.2^2+3*0.2-1 = -.280
0.3*cos0.3-2*0.3^2+3*0.3-1 = 0.019
In order for the function to go from y=-.280 to y=0.019 on the interval [0.2,0.3], it must pass through y = 0 at least once.
-x*sinx -4x + 3 =0
Evaluate the function at 0.2 and 0.3:
0.2*cos0.2-2*0.2^2+3*0.2-1 = -.280
0.3*cos0.3-2*0.3^2+3*0.3-1 = 0.019
In order for the function to go from y=-.280 to y=0.019 on the interval [0.2,0.3], it must pass through y = 0 at least once.
-x*sinx -4x + 3 =0