If the function f(x)=x^3+2 p x^2+27 x+16 is s | vedclass

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(C) Given the function $f(x) = x^3 + 2px^2 + 27x + 16$. For $f(x)$ to be strictly increasing for all $x \in R$,we must have $f'(x) > 0$ for all $x \in

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$\left(\frac{-9}{2}, \frac{9}{2}\right)$

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(C) Given the function $f(x) = x^3 + 2px^2 + 27x + 16$. For $f(x)$ to be strictly increasing for all $x \in R$,we must have $f'(x) > 0$ for all $x \in R$. First,find the derivative: $f'(x) = 3x^2 + 4px + 27$. Since $f'(x)$ is a quadratic expression with a positive leading coefficient $(3 > 0)$,it will be strictly positive for all $x$ if and only if its discriminant $D The discriminant $D$ is given by $D = (4p)^2 - 4(3)(27)$. Setting $D $16p^2 - 324 $p^2 - \frac{324}{16} $p^2 - \frac{81}{4} $(p - \frac{9}{2})(p + \frac{9}{2}) Thus,the range of $p$ is $p \in \left(-\frac{9}{2}, \frac{9}{2}\right)$.

If the function $f(x)=x^3+2 p x^2+27 x+16$ is strictly increasing for all $x \in R$,then the range of $p$ is

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