定义域 f(x)=-sqrt(8ax+1)

解答

定义域

f (x) = - 8 a x + 1

x \geq - \frac{1}{8 a}

求解步骤

对于根式找到非负值:

x \geq - \frac{1}{8 a}

函数定义域

x \geq - \frac{1}{8 a}

d o main f (x) = \frac{7 x}{6 x - 5}

d o main f (x) = g (x) = - 3

d o main - 2 + 4 cos (x)

d o main \frac{( 8 x + 5 )}{2 - 5 x}

d o main \frac{x + 6}{5 - x}