Mathematics - Application of Derivative Question with Solution | TestHub

Consider the function, $f(x)=3x(x-3{)}^2-40$ $\mathrm{Now}S=\left\{x\in \infty k:x^2+30\le 11x\right\}$ So $x^2-11x+30\le 0\Rightarrow x\circ \in [5,6]$ $\therefore f(x)$ will have maximum value for $x=6$ The maximum value of function is, $f(6)=3\times 6\times 3\times 3-40=122$