7.1.4 Exercises

Using the difference quotient, calculate the derivative of

f : ℝ \to ℝ

x \to f (x) : = 4 - x^{2}

at the points

x_{1} = - 2

and

x_{2} = 1

.
Answer:

The difference quotient of $f$ at the point $x_{1} = - 2$ is
and has for $x \to - 2$ the limit $f' (- 2) =$ .
The difference quotient of $f$ at the point $x_{2} = 1$ is
and has for $x \to 1$ the limit $f' (1) =$ .

At the point $x_{1} = - 2$ , we have for the difference quotient

$\frac{Δ (f)}{Δ (x)} = \frac{f (x) - f (x_{1})}{x - x_{1}} = \frac{f (x) - f (- 2)}{x - (- 2)} = \frac{4 - x^{2} - 0}{x + 2} = \frac{(2 - x) (2 + x)}{2 + x} = 2 - x .$
For $x \to x_{1}$ , i.e. for $x \to - 2$ , this difference quotient tends to $2 - (- 2) = 4$ ; hence, $f' (- 2) = 4$ .
At the point $x_{2} = 1$ , we have for the difference quotient

$\frac{Δ (f)}{Δ (x)} = \frac{f (x) - f (x_{2})}{x - x_{2}} = \frac{f (x) - f (1)}{x - 1} = \frac{4 - x^{2} - 3}{x - 1} = \frac{1 - x^{2}}{x - 1} = - \frac{(x - 1) (x + 1)}{x - 1} = - x - 1 .$
For $x \to x_{2}$ , i.e. for $x \to 1$ , this difference quotient has the limit $- 1 - 1 = - 2$ ; hence, $f' (1) = - 2$ .

Explain why the functions

are not differentiable.
Answer:

The derivative of the function $f$ at the point $x_{0} = - 3$ does not exist since the difference quotient
does not converge for $h \to 0$ .
The derivative of the function $g$ at the point $x_{0} = 5$ does not exist since the difference quotient for $h < 0$ has the value
and for $h > 0$ has the value
. Thus, the limit for $h \to 0$ does not exist.

The difference quotient of the function $f$ at the point $x_{0} = - 3$ is

$\frac{Δ (f)}{Δ (x)} = \frac{f (x_{0} + h) - f (x_{0})}{h} = \frac{\sqrt{- 3 + h + 3} - \sqrt{- 3 + 3}}{h} = \frac{\sqrt{h} - 0}{h} = \frac{1}{\sqrt{h}} .$
For $h \to 0$ ( $h > 0$ ), this difference quotient increases infinitely, i.e. the limit of the difference quotient does not exist.
The difference quotient of the function $g$ at the point $x_{0} = 5$ is

$\frac{Δ (g)}{Δ (x)} = \frac{g (x_{0} + h) - g (x_{0})}{h} = \frac{6 \cdot | 2 (5 + h) - 10 | - 6 \cdot | 2 \cdot 5 - 10 |}{h} = \frac{12 | h | - 0}{h} = \frac{12 | h |}{h} .$
For $h < 0$ , since $| h | = - h$ , the difference quotient has the value $- 12$ . In contrast, for $h > 0$ , since $| h | = h$ , it has the value $12$ . Thus, the limit of the difference quotient does not exist. (The limit has always to be unique.)