f^{''}(x)=x^3-4x+1,f^'(0)=2,f(1)=-2

Lösung

f^{^{''}} (x) = x^{3} - 4 x + 1, f^{^{'}} (0) = 2, f (1) = - 2

f (x) = \frac{x ^{5}}{20} - \frac{2 x ^{3}}{3} + \frac{x ^{2}}{2} + 2 x - \frac{233}{60}

Schritte zur Lösung

f^{''} (x) = x^{3} - 4 x + 1

Löse

f^{''} (x) = x^{3} - 4 x + 1 : f^{'} (x) = \frac{x ^{4}}{4} - 2 x^{2} + x + c_{1}

Löse

f^{'} (x) = \frac{x ^{4}}{4} - 2 x^{2} + x + c_{1} : f (x) = \frac{x ^{5}}{20} - \frac{2 x ^{3}}{3} + \frac{x ^{2}}{2} + c_{1} x + c_{2}

Wende die Anfangsbedingung an:

f (x) = \frac{x ^{5}}{20} - \frac{2 x ^{3}}{3} + \frac{x ^{2}}{2} + 2 x - \frac{233}{60}

f (x) = \frac{x ^{5}}{20} - \frac{2 x ^{3}}{3} + \frac{x ^{2}}{2} + 2 x - \frac{233}{60}

x \to 2 lim (x^{2} - 5 x + 6 x^{2} - 4)

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