Страница 270 номер 920, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Найти точки экстремума и значения функции в этих точках: 1) \(y = \frac{(2 - x)^3}{(3 - x)^2}\); 2) \(y = \frac{x^3 + 2x^2}{(x - 1)^2}\); 3) \(y = (x - 1)e^{3x}\); 4) \(y = \sin x + \frac{1}{2}\sin 2x\); 5) \(y = e^{\sqrt{3 - x^2}}\); 6) \(y = \sqrt{e^x - x}\).
1) \(y = \frac{(2 - x)^{3}}{(3 - x)^{2}}\)
\(y'(x) = \dfrac{-3(2 - x)^{2}(3 - x)^{2} + 2(2 - x)^{3}(3 - x)}{(3 - x)^{4}} =\)
\(= \dfrac{(2 - x)^{2}(3 - x)(-9 + 3x + 4 - 2x)}{(3 - x)^{4}} =\)
\(= \dfrac{(2 - x)^{2}(3 - x)(x - 5)}{(3 - x)^{4}} = \dfrac{(2 - x)^{2}(x - 5)}{(3 - x)^{3}}\)
\(\dfrac{(2 - x)^{2}(x - 5)}{(3 - x)^{3}} = 0, x_{1} = 2, x_{2} = 5.\)
\(y'(x) < 0\) при \(x < 2\), \(2 < x < 3\), \(x > 5\);
\(y'(x) > 0\) при \(3 < x < 5\);
\(x = 2\) – стационарная точка, не является точкой экстремума; \(x = 5\) – точка максимума;
\(y(5) = \dfrac{(2 - 5)^{3}}{(3 - 5)^{2}} = -\dfrac{27}{4};\)
2) \(y = \frac{x^{3} + 2x^{2}}{(x - 1)^{2}}\)
\(y'(x) = \dfrac{(3x^{2} + 4x)(x - 1)^{2} - 2(x - 1)(x^{3} + 2x^{2})}{(x - 1)^{4}} =\)
\(= \dfrac{(x - 1)(3x^{3} - 3x^{2} + 4x^{2} - 4x - 2x^{3} - 4x^{2})}{(x - 1)^{4}} =\)
\(= \dfrac{x^{3} - 3x^{2} - 4x}{(x - 1)^{3}};\)
\(x^{3} - 3x^{2} - 4x = 0\)
\(x(x^{2} - 3x - 4) = 0\)
\(x_{1} = 0\) – точка минимума,
\(x_{2{,}3} = \frac{3 \pm \sqrt{9 + 16}}{2}, x_{2} = 4\) – точка минимума,
\(x_{3} = -1\) – точка максимума;
\(y'(x) > 0\) при \(x < -1\), \(0 < x < 1\), \(x > 4\);
\(y'(x) < 0\) при \(-1 < x < 0\), \(1 < x < 4\);
\(y(-1) = \dfrac{-1 + 2}{4} = \dfrac{1}{4}\)
\(y(0) = \dfrac{0}{1} = 0\)
\(y(4) = \dfrac{64 + 32}{9} = \dfrac{96}{9} = \dfrac{32}{3};\)
3) \(y = (x - 1)e^{3x}\)
\(y'(x) = e^{3x} + 3(x - 1)e^{3x} = e^{3x}(1 + 3x - 3) = e^{3x}(3x - 2),\)
\(e^{3x}(3x - 2) = 0\)
\(3x - 2 = 0\)
\(3x = 2, x = \frac{2}{3}\) – точка минимума: \(y'(x) < 0\) при \(x < \frac{2}{3}\), \(y'(x) > 0\) при \(x > \frac{2}{3}\),
\(y\left(\dfrac{2}{3}\right) = \left(\dfrac{2}{3} - 1\right)e^{2} = -\dfrac{e^{2}}{3};\)
4) \(y = \sin x + \frac{1}{2}\sin 2x\)
\(y'(x) = \cos x + \cos 2x\)
\(\cos x + \cos 2x = 0\)
\(\cos x + \cos^{2} x - \sin^{2} x = 0\)
\(\cos x + 2\cos^{2} x - 1 = 0\)
\(\cos x = z, 2z^{2} + z - 1 = 0\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 8}}{4}, z_{1} = \dfrac{1}{2}, z_{2} = -1;\)
\(y'(x) = (2\cos x - 1)(\cos x + 1)\),
\(y'(x) > 0\) при \(-\frac{\pi}{3} + 2\pi n < x < \frac{\pi}{3} + 2\pi n\);
\(y'(x) < 0\) при \(\frac{\pi}{3} + 2\pi n < x < \pi + 2\pi n\)
и при \(\pi + 2\pi n < x < \frac{5\pi}{3} + 2\pi n\);
\(x_{1} = \frac{\pi}{3} + 2\pi n\) – точка максимума; \(x_{2} = \pi + 2\pi n\) – не является точкой экстремума; \(x_{3} = \frac{5\pi}{3} + 2\pi n\) – точка минимума, \(n \in Z\);
\(y\left(\dfrac{\pi}{3} + 2\pi n\right) = \sin\left(\dfrac{\pi}{3} + 2\pi n\right) + \dfrac{1}{2}\sin\left(\dfrac{2\pi}{3} + 4\pi n\right) =\)
\(= \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{4},\)
\(y\left(\dfrac{5\pi}{3} + 2\pi n\right) = \sin\left(\dfrac{5\pi}{3} + 2\pi n\right) + \dfrac{1}{2}\sin\left(\dfrac{10\pi}{3} + 4\pi n\right) =\)
\(= -\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}\dfrac{\sqrt{3}}{2} = -\dfrac{3\sqrt{3}}{4};\)
5) \(y = e^{\sqrt{3 - x^{2}}}\)
\(y'(x) = e^{\sqrt{3 - x^{2}}} \cdot (-2x) \cdot \frac{1}{2\sqrt{3 - x^{2}}} = -\frac{xe^{\sqrt{3 - x^{2}}}}{\sqrt{3 - x^{2}}},\)
\(-\frac{x}{\sqrt{3 - x^{2}}} = 0, x = 0\) – точка максимума:
\(y'(x) > 0\) при \(-\sqrt{3} < x < 0\), \(y'(x) < 0\) при \(0 < x < \sqrt{3}\);
\(y(0) = e^{\sqrt{3}}\).
6) \(y = \sqrt{e^{x} - x}\)
\(y'(x) = \dfrac{1}{2\sqrt{e^{x} - x}} \cdot (e^{x} - 1)\)
\(\dfrac{e^{x} - 1}{2\sqrt{e^{x} - x}} = 0\)
\(e^{x} - 1 = 0\)
\(e^{x} = 1, x = 0\) – точка минимума: \(y'(x) < 0\) при \(x < 0\), \(y'(x) > 0\) при \(x > 0\);
\(y(0) = \sqrt{e^{0} - 0} = 1.\)
