6
$\begingroup$

I need to find the slope at a=5, using the definition for the function $f(x)=\sqrt{x^2 -9}$,

$$f'(x) = \lim_{\Delta x \to 0} {f(x+\Delta x)\over \Delta x}$$

The answer book says the slope is ${1\over 4}$

Here's what I did,

$$f'(x) = \lim_{\Delta x \to 0} {(\sqrt{(x+\Delta x)^2 -9} - \sqrt {x^2 -9} )(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)})\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}} (1)\\=\lim_{\Delta x \to 0} {(x+\Delta x)^2 -9 -x^2 +9\over \Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}} (2)\\=\lim_{\Delta x \to 0}{x^2 +2x \Delta x+ \Delta x^2 -x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}} (3)\\=\lim_{\Delta x \to 0}{2x\Delta x +\Delta x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}} (4)\\=\lim_{\Delta x \to 0} {2x+\Delta x\over(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}} (5)\\={2x\over \sqrt{x^2-9+x^2-9}} (6)\\={2x\over2 \sqrt{x^2 -9}} (7)\\={x\over \sqrt {x^2 -9}} (8)$$

Now I substitute 5, and I don't get 1/4!!

What have I done wrong??

Thanks

$\endgroup$
  • $\begingroup$ You have asked several similar questions already. Let's take a break. In every question, it's because you made a simply arithmetic error. Perhaps explore one answer and work your other problems more carefully before asking yet another almost identical question. $\endgroup$ – Emily Jan 9 '15 at 16:58
  • $\begingroup$ Your calculation was fine, and then there were a couple of algebra errors in a row. As a minor suggestion, since we are interested in $x=5$, it might have been a good idea not to proceed "generally," but instead to use $x=5$ from the beginning. $\endgroup$ – André Nicolas Jan 9 '15 at 17:00
  • $\begingroup$ @Arkamis I don't see a problem with the user asking the questions he/she did. If it bothers you that their mistakes are arithmetic, maybe you should move on. Unless the user is violating StackExchange rules, it isn't for any of us to tell them what questions they can and can't ask. $\endgroup$ – layman Jan 9 '15 at 17:02
  • $\begingroup$ @user46944 It is generally frowned-upon on this site to rapidly and serially ask questions that are abstract duplicates. $\endgroup$ – Emily Jan 9 '15 at 17:03
  • 1
    $\begingroup$ (Cont.) So there were errors. But in fact the slope is not $\frac{1}{4}$, so there is also an error in the answer file. $\endgroup$ – André Nicolas Jan 9 '15 at 17:05
2
$\begingroup$

(5)to (6) step !!!you have error $$\sqrt{(x+\Delta x)^2-9}+\sqrt{(x)^2-9} \neq \sqrt{(x)^2-9+(x)^2-9} $$in fact $$\sqrt{a+b} \neq \sqrt{a} +\sqrt{b}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

In step 6, the denominator should be $2 \sqrt{x^2-9}$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Also, are you sure the book's answer is $\frac{1}{4}$? It should be $\frac{5}{4}$. $\endgroup$ – KittyL Jan 9 '15 at 16:58
  • $\begingroup$ Well, the book had couple mistakes in the answer book before... $\endgroup$ – didgocks Jan 9 '15 at 16:58
  • $\begingroup$ $\sqrt{x^2-9}+\sqrt{x^2-9}=2\sqrt{x^2-9}$. $\endgroup$ – KittyL Jan 9 '15 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.