6. Application of derivatives — Maths STD 12 Science — Question

continuous function.

putting derivative $=0,$ we get

$\left(4 \times-1 \times\{7-x\}^{3} \times\{2+x\}^{5}\right)+\left(5 \times\{7-x\}^{4}\{2+x\}^{4}\right)=0$

or $, 5\{7-x\}=4\{2+x\}$ since $x \in(-2,7)$ so $\{x+2\} \neq 0$ \& $\{7-x\} \neq 0$

Therefore $9 x=27$ or, $x=3$

Hence the maximum value will be $(7-3)^{4}(2+3)^{5}$ or $8 \times 10^{5}$ or $4^45^5$