F(x)=[x^2]+1
it equals 1
That's a Gaussian distribution.
14x
Are you trying to solve for x? Fx = x2 - 3 x2 - Fx - 3 = 0 x2 - Fx = 3 x2 - Fx + (F/2)2 = 3 + (F/2)2 (x - F/2)2 = 3 + (F/2)2 x - F/2 = ±[ 3 + (F/2)2 ]1/2 x = F/2 ± [ 3 + (F/2)2 ]1/2
yup
F(x)=[x^2]+1
it equals 1
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
That's a Gaussian distribution.
f^2 + 2f = f (f + 2)
14x
-2, 1.74 and 0.46
Are you trying to solve for x? Fx = x2 - 3 x2 - Fx - 3 = 0 x2 - Fx = 3 x2 - Fx + (F/2)2 = 3 + (F/2)2 (x - F/2)2 = 3 + (F/2)2 x - F/2 = ±[ 3 + (F/2)2 ]1/2 x = F/2 ± [ 3 + (F/2)2 ]1/2
Let f ( x ) = 3 x 5 and g ( x ) = 3 x 2 4 x
f(x)=x2-x3 f(2) = 4-8 = -4 f'(x) =2x-3x2 f'(2) = 4-12=-8 f''(x) = 2 -6x f''(2)= -10 f'''(x)= -6 f(n)(x) = 0 for all n > 3. f(x) = f(2) + (x-2) f'(2) / 1! + (x-2)2 f''(2) /2! + (x-2)3 f'''(2)/3! + . . . f(x) = -4 -8(x-2) -10(x-2)22/2-6(x-2)3/6 + 0 + 0 + ... f(x) = -4 -8(x-2) -5(x-2)2 - (x-2)3
If: f/7 = 8 Then: f = 56