Let a polynomial f(x) of degree 4 satisfy f(0)=10 , f'(0)=36 , f''(0)=60 , f'''(0)=-120 , and f''''(0)=72 . If M is the local maximum value of the function f(x) , then M is equal to

Step-by-step solution

Using Maclaurin's series, the polynomial f(x) can be written as: f(x) = f(0) + f'(0)x + f''(0) 2! x^2 + f'''(0) 3! x^3 + f''''(0) 4! x^4 Substituting the given values: f(x) = 10 + 36x + 60 2 x^2 - 120 6 x^3 + 72 24 x^4 f(x) = 3x^4 - 20x^3 + 30x^2 + 36x + 10 Differentiating with respect to x : f'(x) = 12x^3 - 60x^2 + 60x + 36 = 12(x^3 - 5x^2 + 5x + 3) For critical points, f'(x) = 0 . By inspection, x = 3 is a root. 12(x - 3)(x^2 - 2x - 1) = 0 Equating the quadratic factor to zero, the roots are x = 2 4 - 4(-1) 2 = 1