Let the function f(x)=2 x^3+(2 p-7) x^2+3(2 p | vedclass

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(C) Given the function $f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6$. First,find the derivative $f'(x) = 6x^2 + 2(2p-7)x + 3(2p-9)$. For the function to have

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

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(C) Given the function $f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6$. First,find the derivative $f'(x) = 6x^2 + 2(2p-7)x + 3(2p-9)$. For the function to have a local maximum at $x  0$,the quadratic equation $f'(x) = 0$ must have two distinct real roots,one negative and one positive. Let the roots be $\alpha$ and $\beta$ such that $\alpha For a quadratic $ax^2 + bx + c = 0$ to have roots of opposite signs,the product of the roots must be negative,i.e.,$\frac{c}{a} Here,$a = 6$ and $c = 3(2p-9)$. Thus,$\frac{3(2p-9)}{6} This implies $2p - 9 Therefore,the set of all values of $p$ is $\left(-\infty, \frac{9}{2}\right)$.

Let the function $f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6$ have a maxima for some value of $x < 0$ and a minima for some value of $x > 0$. Then,the set of all values of $p$ is $......$

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