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Calculus Level pending

Consider the real function f ( x ) = 3 x 4 + 2 x 3 + 11 x 2 − 40 x + 24 1 − x f(x) = \dfrac{3x^4 + 2x^3 + 11x^2 - 40 x + 24}{\sqrt{1-x}} .

Evaluate f ( 1 ∫ 0 1 f ′ ′ ( x ) ∂ x ) f \left ( \dfrac{1}{ \int_{0}^{1} f''(x) \partial{x} } \right ) to the nearest integer.


The answer is 23.

Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

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1 solution

Taking the first derivative of f ( x ) f(x) , we have ∂ ∂ x ( 3 x 4 + 2 x 3 + 11 x 2 − 40 x + 24 1 − x ) = − 7 2 ( 3 x 2 + 4 x + 8 ) 1 − x \dfrac{\partial}{\partial{x}} \left ( \dfrac{3x^4 + 2x^3 + 11x^2 - 40 x + 24}{\sqrt{1-x}} \right ) = -\dfrac{7}{2} (3x^2+4x+8) \displaystyle \sqrt{1-x} .

Since ∫ 0 1 f ′ ′ ( x ) ∂ x = f ′ ( 1 ) − f ′ ( 0 ) \displaystyle \int_{0}^{1} f''(x) \partial{x} = f'(1) - f'(0) , we have ∫ 0 1 f ′ ′ ( x ) ∂ x = 28 \displaystyle \int_{0}^{1} f''(x) \partial{x} = 28 .

Finally, direct substitution yields f ( 1 28 ) = 1542483 21 307328 ≈ 23.0000036 ≈ 23. f \left ( \dfrac{1}{28} \right ) = \dfrac{1542483 \sqrt{21}}{307328} \approx 23.0000036 \approx \boxed{23.}

PS: Look how close to the integer this result is!

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