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To find f(-3), we substitute -3 into the function f(x) = x^2 + x: f(-3) = (-3)^2 + (-3) = 9 - 3 = 6
If f(x) = 35/5 + 3 then its inverse is f(x) = 5/3*(x - 3).
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
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
f(x) = x2 + 3x - 2 then f'(x) = 2x + 3 and then then f'(2) = 2*2 + 3 = 4+3 = 7