2020-수학-가형-홀수-11¶
문제¶
풀이¶
from sympy import *
a, x = symbols('a x')
first_der = diff(a*x**2-2*sin(2*x),x)
second_der = diff(first_der,x)
second_der
\[\displaystyle 2 a + 8 \sin{\left(2 x \right)}\]
count = 0
for n in range(-10,10):
fxx = second_der.subs(a,n)
fxx_values = [fxx.subs(x,k/10).evalf() for k in range(-100,100)]
interval = Interval(min(fxx_values), max(fxx_values))
print([n, interval.contains(0), interval])
if interval.contains(0) :
count +=1
count
[-10, False, Interval(-27.9999216524056, -12.0000783475944)]
[-9, False, Interval(-25.9999216524056, -10.0000783475944)]
[-8, False, Interval(-23.9999216524056, -8.00007834759437)]
[-7, False, Interval(-21.9999216524056, -6.00007834759437)]
[-6, False, Interval(-19.9999216524056, -4.00007834759437)]
[-5, False, Interval(-17.9999216524056, -2.00007834759437)]
[-4, False, Interval(-15.9999216524056, -0.0000783475943721967)]
[-3, True, Interval(-13.9999216524056, 1.99992165240563)]
[-2, True, Interval(-11.9999216524056, 3.99992165240563)]
[-1, True, Interval(-9.99992165240563, 5.99992165240563)]
[0, True, Interval(-7.99992165240563, 7.99992165240563)]
[1, True, Interval(-5.99992165240563, 9.99992165240563)]
[2, True, Interval(-3.99992165240563, 11.9999216524056)]
[3, True, Interval(-1.99992165240563, 13.9999216524056)]
[4, False, Interval(0.0000783475943721967, 15.9999216524056)]
[5, False, Interval(2.00007834759437, 17.9999216524056)]
[6, False, Interval(4.00007834759437, 19.9999216524056)]
[7, False, Interval(6.00007834759437, 21.9999216524056)]
[8, False, Interval(8.00007834759437, 23.9999216524056)]
[9, False, Interval(10.0000783475944, 25.9999216524056)]
7