1. ## Limits

Find:
a) the slope of the tangent
b) the slope at $x = 7$
c) where the curve is differentiable

equation:
$f(x) = \sqrt (x - 3)$

a)

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

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

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

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

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

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

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

Textbook Answer: $m = \frac{1}{(x + 3)^2}$

Are questions b and c correct?

b)

$y = \frac{-1}{(x + 3)^2}x + b$
$y = \frac{-1}{(7 + 3)^2} \times 7 + b$
$y = \frac{-1}{10^2} \times 7 + b$
$y = \frac{-7}{100} + b$

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

c)

$x ER$ (x at all real numbers)

2. Originally Posted by Macleef
Find:
a) the slope of the tangent
b) the slope at $x = 7$
c) where the curve is differentiable

a)

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

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

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

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

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

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

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

Textbook Answer: $m = \frac{1}{(x + 3)^2}$

Are questions b and c correct?

b)

$y = \frac{-1}{(x + 3)^2}x + b$
$y = \frac{-1}{(7 + 3)^2} \times 7 + b$
$y = \frac{-1}{10^2} \times 7 + b$
$y = \frac{-7}{100} + b$

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

c)

$x ER$ (x at all real numbers)
For a) your problem is that you are correct. The book is wrong.

b) is also correct.

c) What happens when x = -3?

-Dan

3. Originally Posted by Macleef
Find:
a) the slope of the tangent
b) the slope at $x = 7$
c) where the curve is differentiable

a)

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

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

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

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

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

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

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

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

Are questions b and c correct? Mr F says: b) is wrong. You're meant to substitute x = 7 into ${\color{red}m = \frac{-1}{(x + 3)^2}}$.

And c) is wrong. You're meant to consider when ${\color{red}m = \frac{-1}{(x + 3)^2}}$ 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)
$y = \frac{-1}{(x + 3)^2}x + b$
$y = \frac{-1}{(7 + 3)^2} \times 7 + b$
$y = \frac{-1}{10^2} \times 7 + b$
$y = \frac{-7}{100} + b$

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

c)

$x ER$ (x at all real numbers)
..

4. So 3 is the limit ?

5. Originally Posted by Macleef
So 3 is the limit ?
What do you mean? Limit of what?

6. Originally Posted by mr fantastic
What do you mean? Limit of what?
of the curve

7. Originally Posted by Macleef
of the curve
I have no idea what you're asking ....

"So 3 is the limit of the curve" makes absolutely no sense. State the complete question that you think 3 is the answer to.

8. is the curve differentiable at x = 3 because the restriction is at this x value? or is f(x) not differentiable because it's a discontinuity curve?

9. Originally Posted by Macleef
is the curve differentiable at x = 3 because the restriction is at this x value? or is f(x) not differentiable because it's a discontinuity curve?
Of course it is. $m = \frac{-1}{(x + 3)^2}$ has no trouble with x = 3: m = -1/36, a perfectly fine slope.

As both I and topsquark have already said, $m = \frac{-1}{(x + 3)^2}$ has trouble when x = -3. Because the denominator is equal to zero and hence the expression becomes undefined.

So clearly the curve is differentiable for all real values of x except x = -3.

10. Originally Posted by topsquark
For a) your problem is that you are correct. The book is wrong.

b) is also correct. Mr F begs to differ....

c) What happens when x = -3?

-Dan
Nope. I nearly thought so too at first glance .....