函数(x)=(x-5)(x)^dfrac (2{3)}的拐点是________.

步骤 1：求一阶导数
首先，我们需要求出函数$f(x)=(x-5){x}^{\dfrac {2}{3}}$的一阶导数$f'(x)$。使用乘积法则和幂函数的导数规则，我们得到：
$$f'(x) = (x-5)'\cdot{x}^{\dfrac {2}{3}} + (x-5)\cdot({x}^{\dfrac {2}{3}})'$$
$$f'(x) = {x}^{\dfrac {2}{3}} + (x-5)\cdot\dfrac {2}{3}{x}^{-\dfrac {1}{3}}$$
$$f'(x) = {x}^{\dfrac {2}{3}} + \dfrac {2}{3}(x-5){x}^{-\dfrac {1}{3}}$$
$$f'(x) = {x}^{\dfrac {2}{3}} + \dfrac {2}{3}x^{\dfrac {2}{3}} - \dfrac {10}{3}x^{-\dfrac {1}{3}}$$
$$f'(x) = \dfrac {5}{3}x^{\dfrac {2}{3}} - \dfrac {10}{3}x^{-\dfrac {1}{3}}$$
步骤 2：求二阶导数
接下来，我们需要求出函数$f(x)$的二阶导数$f''(x)$。对$f'(x)$求导，我们得到：
$$f''(x) = \dfrac {5}{3}\cdot\dfrac {2}{3}x^{-\dfrac {1}{3}} + \dfrac {10}{3}\cdot\dfrac {1}{3}x^{-\dfrac {4}{3}}$$
$$f''(x) = \dfrac {10}{9}x^{-\dfrac {1}{3}} + \dfrac {10}{9}x^{-\dfrac {4}{3}}$$
$$f''(x) = \dfrac {10}{9}x^{-\dfrac {1}{3}}(1 + x^{-1})$$
步骤 3：求拐点
拐点是二阶导数$f''(x)$等于0的点。因此，我们解方程$f''(x) = 0$：
$$\dfrac {10}{9}x^{-\dfrac {1}{3}}(1 + x^{-1}) = 0$$
由于$\dfrac {10}{9}x^{-\dfrac {1}{3}}$不等于0，我们只需解$1 + x^{-1} = 0$：
$$1 + x^{-1} = 0$$
$$x^{-1} = -1$$
$$x = -1$$
因此，拐点的横坐标是$x = -1$。将$x = -1$代入原函数$f(x)$，我们得到拐点的纵坐标：
$$f(-1) = (-1-5)(-1)^{\dfrac {2}{3}}$$
$$f(-1) = -6\cdot1$$
$$f(-1) = -6$$