![y = log_3(11 + 4x - x^2) - 2\\\\ (f(g(x)))' = \frac{dg(x)}{dx}\frac{df(g(x))}{dg(x)} = g'(x)\frac{df(g(x))}{dg(x)}\\\\ y' = (4 - 2x) \frac{1}{(11 + 4x - x^2)ln(3)}\\\\ (4 - 2x) \frac{1}{(11 + 4x - x^2)ln(3)} = 0,\\\\ 11 + 4x - x^2 \ne 0,\\\\ 4 - 2x = 0, -2x = -4, \ \underline{x = 2}\\\\ 11 + 4x - x^2 = 0, \\\\\ D = 16 + 44 = 60\\\\ x_1 = -\frac{-4 - \sqrt{60}}{2} = 2 + \sqrt{15}\\\\ x_2 = -\frac{-4 + \sqrt{60}}{2} = 2 - \sqrt{15}](https://tex.z-dn.net/?f=y+%3D+log_3%2811+%2B+4x+-+x%5E2%29+-+2%5C%5C%5C%5C+%28f%28g%28x%29%29%29%27+%3D+%5Cfrac%7Bdg%28x%29%7D%7Bdx%7D%5Cfrac%7Bdf%28g%28x%29%29%7D%7Bdg%28x%29%7D+%3D+g%27%28x%29%5Cfrac%7Bdf%28g%28x%29%29%7D%7Bdg%28x%29%7D%5C%5C%5C%5C+y%27+%3D+%284+-+2x%29+%5Cfrac%7B1%7D%7B%2811+%2B+4x+-+x%5E2%29ln%283%29%7D%5C%5C%5C%5C+%284+-+2x%29+%5Cfrac%7B1%7D%7B%2811+%2B+4x+-+x%5E2%29ln%283%29%7D+%3D+0%2C%5C%5C%5C%5C+11+%2B+4x+-+x%5E2+%5Cne+0%2C%5C%5C%5C%5C+4+-+2x+%3D+0%2C+-2x+%3D+-4%2C+%5C+%5Cunderline%7Bx+%3D+2%7D%5C%5C%5C%5C+11+%2B+4x+-+x%5E2+%3D+0%2C+%5C%5C%5C%5C%5C+D+%3D+16+%2B+44+%3D+60%5C%5C%5C%5C+x_1+%3D+-%5Cfrac%7B-4+-+%5Csqrt%7B60%7D%7D%7B2%7D+%3D+2+%2B+%5Csqrt%7B15%7D%5C%5C%5C%5C+x_2+%3D+-%5Cfrac%7B-4+%2B+%5Csqrt%7B60%7D%7D%7B2%7D+%3D+2+-+%5Csqrt%7B15%7D+)
Методом интервалов находим, что y' > 0:
![x \in (2 - \sqrt{15}; 2) \cup (2 + \sqrt{15}; +\infty)](https://tex.z-dn.net/?f=x+%5Cin+%282+-+%5Csqrt%7B15%7D%3B+2%29+%5Ccup+%282+%2B+%5Csqrt%7B15%7D%3B+%2B%5Cinfty%29+)
тогда x = 2 - точка максимума.
![\max\limits_{x} y = y(2) = log_3(11 + 8 - 4) - log_39 = log_3\frac{15}{9} = log_3\frac{5}{3}](https://tex.z-dn.net/?f=%5Cmax%5Climits_%7Bx%7D+y+%3D+y%282%29+%3D+log_3%2811+%2B+8+-+4%29+-+log_39+%3D+log_3%5Cfrac%7B15%7D%7B9%7D+%3D+log_3%5Cfrac%7B5%7D%7B3%7D)
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Точки (х,у) подставляем в уравнение
1=с
-2=а+б+с -> а+б=-3
-1=2а+4б+1 -> 2а+4б=-2 -> а+2б=-1 -> б=2 , а=-5
6 3/8 + 2 5/9 = 51/8 + 23/9 =(51 × 9 + 23 × 8)/72 = 459/72 + 184/72 = 643/72 = 8 67/72