Given f(x) = x^3 if x<1 and ax^2+bx +c if x ≥ 1 find the values of a, b, and c such that f''(1) exists

Answer 1

y = 3x^2 -3x +1 is a quadratic equation that intersects y=x^3 at (1,1) with the same 1st & 2nd derivatives, so that the curves are not discontinuous at (1,1)

so a=3, b =-3, c= 1

y=3x^2 -3x + 1 = 3(x-1/2)^2 + 1/4

the quadratic > cubic for x<1, but

the quadratric < cubic for x>1

at x=1 the quadratic = the cubic and

their derivatives are also equal

Answer 2

f(1) = 1³ = 1, so a + b + c = 1

f'(x) = 3x²; f'(1) = 3 but f'(x) = 2ax + b; f'(1) = 2a + b = 3

f''(x) = 6x; f''(1) = 6; f''(x) = 2a; f''(1) = 2a; 2a = 6; a = 3; b = 3 - 6 = - 3; and c = 1

So,. f(x) = x³ if x < 1 and f(x) = 3x² - 3x + 1 for x ≥ 1