13.2 Continuous of Functions 函数的连续性 - 独中课程 | 线上补习

高二数学 | 高级数学

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13.2 Continuous of Functions 函数的连续性

[1]

Given that f (x) = $x^2 - 3x + \frac{1}{x} - 1$ , find the values of

已知f (x) = $x^2 - 3x + \frac{1}{x} - 1$ ，求值

(a) $\lim\limits_{h \to o} \frac{f(-1+h)-f(-1))}{h}$

(b) $\lim\limits_{h \to o} \frac{f(-1)-f(-1-h)}{h}$

(a) – 6

(b) –6

[2]

Give that g(x) = $\begin{cases} \frac{x^3 - x}{ x^2 + x} & , x < 1 (x \neq 0, -1) \\ 0 & , x = 0 \\ 1 - x & , x \geq 1\end{cases}$

(a)
where is g discontinuous
在那里g 不连续

(b)
where is g not differentiable
在那里g 不可微

(a) 0, –1

(b) 0, –1, 1

[3]

If g(x) = $\begin{cases} \frac{1-\sqrt{1-x}}{x} & , x < 0 \\ a + bx & , x \geq 0 \end{cases}$ , , if f is differentiable at x = 0, find a and b.

$\frac{1}{2}$ , $\frac{1}{8}$