Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^2y}{dx^2}\) for the function \(y=x^3+\dfrac{4}{x^2}\).
\(\begin{aligned} y&=x^3+\dfrac{4}{x^2} \\ &=x^3+4x^{-2} \\\\ \dfrac{dy}{dx}&=3x^2-8x^{-3} \\ \dfrac{dy}{dx}&=3x^2-\dfrac{8}{x^3} \\\\ \dfrac{d^2y}{dx^2}&=6x+24x^{-4} \\ \dfrac{d^2y}{dx^2}&=6x+\dfrac{24}{x^4}. \end{aligned}\)
If \(g(x)=2x^3+3x^2-7x-9\), find \(g''\left( \dfrac{1}{4} \right)\) and \(g''(-1)\).
\(\begin{aligned} g(x)&=2x^3+3x^2-7x-9 \\ g'(x)&=6x^2+6x-7 \\ g''(x)&=12x+6. \end{aligned}\)
Thus,
\(\begin{aligned} g''\left( \dfrac{1}{4} \right)&=12\left( \dfrac{1}{4} \right)+6 \\ &=3+6 \\ &=9. \end{aligned}\)
\(\begin{aligned} g''(-1)&=12(-1)+6 \\ &=-12+6 \\ &=-6. \end{aligned}\)
Given the function \(f(x)=x^3+2x^2+3x+4\), find the values of \(x\) such that \(f'(x)=f''(x)\).
Given \(f(x)=x^3+2x^2+3x+4\).
Then,
\(f'(x)=3x^2+4x+3\), \(f''(x)=6x+4\).
For \(f'(x)=f''(x)\),
\(\begin{aligned} 3x^2+4x+3&=6x+4 \\ 3x^2-2x-1&=0 \\ (3x+1)(x-1)&=0 \end{aligned}\)
\(x=-\dfrac{1}{3}\) or \(x=1\).
Therefore, the values of \(x\) are \(-\dfrac{1}{3}\) and \(1\).
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