Question

# The maximum value of the function $$f(x)=2x^3-15x^2+36x-48$$ on the set $$A=\{x|x^2+20\leq 9x\}$$ is

Solution

## $$x² + 20≤ 9x$$$$x² -9x +20 ≤ 0$$$$x² -4x -5x +20 ≤ 0$$$$x( x -4) -5(x -4) ≤ 0$$$$(x -4)(x -5) ≤ 0$$$$4 ≤ x ≤ 5$$$$\mathrm{A}=\{4\leq \mathrm{x}\leq 5\}$$now,$$f(x ) = 2x³ - 15x² +36x -48$$differentiate wrt $$x$$$$f'(x) = 6x² -30x +36 = 6(x²-5x +6)$$$$= 6(x -2)(x -3)$$So $$\mathrm{f}(\mathrm{x})$$ is increasing in $$(-\infty,2)\cup (3, \infty)$$maximum value of $$f(x)$$ at $$x = 5$$$$f(5) = 2(5)³-15(5)² +36(5) -48$$$$= 250 -375 + 180 -48$$$$= 430 - 423 = 7$$$$\therefore \mathrm{f}_{\max}=\mathrm{f}(5)=7$$Mathematics

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