**Find:** **a) the slope of the tangent ** **b) the slope at $\displaystyle x = 7$** **c) where the curve is differentiable** **a)**
$\displaystyle m = \lim_{h \to \-0} \frac{f(x + h) - f(x)}{h} $

$\displaystyle m = \lim_{h \to \-0} \frac{\frac {1}{x + h + 3} - \frac {1}{x + 3}}{h} $

$\displaystyle m = \lim_{h \to \-0} \frac{\frac {x + 3 - x - h - 3}{(x + h + 3)(x + 3)}}{h} $

$\displaystyle m = \lim_{h \to \-0} \frac{-h}{(x + h + 3)(x + 3)} \times \frac {1}{h} $

$\displaystyle m = \lim_{h \to \-0} \frac{-1}{(x + h + 3)(x + 3)} $

$\displaystyle m = \frac{-1}{(x + 0 + 3)(x + 3)} $

$\displaystyle m = \frac{-1}{(x + 3)^2} $

Mr F says: CORRECT. **Textbook Answer:** $\displaystyle m = \frac{1}{(x + 3)^2} $

Mr F says: INCORRECT. *I get a negative answer instead of a positive answer, I can't seem to find my mistake, please help?* *Are questions b and c correct? Mr F says: b) is wrong. You're meant to substitute x = 7 into *$\displaystyle {\color{red}m = \frac{-1}{(x + 3)^2}} $. And c) is wrong. You're meant to consider when $\displaystyle {\color{red}m = \frac{-1}{(x + 3)^2}} $ is defined. In this case, it's easier to consider when it's NOT defined. It doesn't look to healthy at the real number x = - 3 ..... **b) **
$\displaystyle y = \frac{-1}{(x + 3)^2}x + b $

$\displaystyle y = \frac{-1}{(7 + 3)^2} \times 7 + b $

$\displaystyle y = \frac{-1}{10^2} \times 7 + b $

$\displaystyle y = \frac{-7}{100} + b $

$\displaystyle m = \frac{-7}{100}$

**c)**
$\displaystyle x ER$ (x at all real numbers)