What is the solution for f^{''}(x)=9x^3+4x^2+7x ?

The solution for f^{''}(x)=9x^3+4x^2+7x is f(x)=(9x^5}{20}+\frac{x^4)/3+(7x^3)/6+c_{1}x+c_{2}

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f^{''}(x)=9x^3+4x^2+7x

f^{^{''}} (x) = 9 x^{3} + 4 x^{2} + 7 x

f (x) = \frac{9 x ^{5}}{20} + \frac{x ^{4}}{3} + \frac{7 x ^{3}}{6} + c_{1} x + c_{2}

Solution steps

f^{''} (x) = 9 x^{3} + 4 x^{2} + 7 x

Solve

f^{''} (x) = 9 x^{3} + 4 x^{2} + 7 x : f^{'} (x) = \frac{9 x ^{4}}{4} + \frac{4 x ^{3}}{3} + \frac{7 x ^{2}}{2} + c_{1}

Solve

f^{'} (x) = \frac{9 x ^{4}}{4} + \frac{4 x ^{3}}{3} + \frac{7 x ^{2}}{2} + c_{1} : f (x) = \frac{9 x ^{5}}{20} + \frac{x ^{4}}{3} + \frac{7 x ^{3}}{6} + c_{1} x + c_{2}

f (x) = \frac{9 x ^{5}}{20} + \frac{x ^{4}}{3} + \frac{7 x ^{3}}{6} + c_{1} x + c_{2}

x \to 0 lim (∣ x - 2 ∣ + 3)

\frac{\partial}{\partial y} (sin (y) cos (y))

y = 12 - x^{2}, y = - 3 x - 16

(e^{2 x} \cdot 2)^{^{'}}

y (θ) = θ cos (θ)

What is the solution for f^{''}(x)=9x^3+4x^2+7x ?
The solution for f^{''}(x)=9x^3+4x^2+7x is f(x)=(9x^5}{20}+\frac{x^4)/3+(7x^3)/6+c_{1}x+c_{2}