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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