Q.Find the values of $x$, when the marginal function of $y = x^{3} + 10x^{2} - 48x + 8$ is twice the $x$.

Marginal function $=3x^{2}+20x-48$; setting it $=2x$ gives $x^{2}+6x-16=0\Rightarrow x=2$ or $x=-8$.

In the TN HSC Class-11 Business Mathematics syllabus, the marginal function of $y$ is $\dfrac{dy}{dx}$.

Step 1 — Differentiate.

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(x^{3}+10x^{2}-48x+8\right)=3x^{2}+20x-48.$

Step 2 — Set marginal function equal to $2x$.

$3x^{2}+20x-48=2x.$

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