Calculus Problem

#1 $Old$ November 13th, 2006, 06:48 PM

Find the point(s) on the graph of

where the slope is 5. (Thanks!)

Soroban · #2 $Old$ November 13th, 2006, 07:30 PM

Hello, Affinity!

Quote:

Find the point(s) on the graph of: $f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}}$ where the slope is 5.

We have: . $f(x)\:=\:\frac{x - 3x^{\frac{3}{2}}}{x^{\frac{1}{2}}} \:=\:x^{\frac{1}{2}} - 3x$

Then: . $f'(x)\;=\;\frac{1}{2}x^{-\frac{1}{2}} - 3 \;=\;5$

We have: . $\frac{1}{2\sqrt{x}} \;=\;8\quad\Rightarrow\quad 16\sqrt{x}\;=\;1\quad\Rightarrow\quad \sqrt{x} \,=\,\frac{1}{16}\quad\Rightarrow\quad x\,=\,\frac{1}{256}$

Then: . $f\left(\frac{1}{256}\right)\:=\:\sqrt{\frac{1}{256}} - 3\left(\frac{1}{256}\right) \;=\;\frac{13}{256}$

Answer: . $\left(\frac{1}{256},\,\frac{13}{256}\right)$

srinivas · #3 $Old$ November 13th, 2006, 10:19 PM

helo everybody
siriban u have written a different function i think

srinivas · #4 $Old$ November 13th, 2006, 10:20 PM

sorry thats soroban not siriban!!!

topsquark · #5 $Old$ November 14th, 2006, 04:59 AM

Quote:

Originally Posted by Affinity $View Post$

Find the point(s) on the graph of

where the slope is 5. (Thanks!)

We are looking for x values where the first derivative is equal to 5.

$y = \frac{x^2 + 3x - 1}{x}$

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

$y' = \frac{2x^2 + 3x - x^2 - 3x + 1}{x^2}$

$y' = \frac{x^2 + 1}{x^2}$

So:
$5 = \frac{x^2 + 1}{x^2}$

$5x^2 = x^2 + 1$

$4x^2 - 1 = 0$

$(2x + 1)(2x - 1) = 0$

Thus $x = \pm 1/2$

-Dan

Soroban · #6 $Old$ November 14th, 2006, 06:43 AM

Help!
Am I the only one seeing: . $\boxed{f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}}}$ ??

topsquark · #7 $Old$ November 14th, 2006, 12:41 PM

Quote:

Originally Posted by Soroban $View Post$

Help!
Am I the only one seeing: . $\boxed{f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}}}$ ??

Interesting. He made an earlier post here with that function. Now as I look at the other post all I can see is the function for this one.

An idea: perhaps both of us need to clear our browser's cache?

-Dan

earboth · #8 $Old$ November 14th, 2006, 12:54 PM

Quote:

Originally Posted by Affinity $View Post$

Find the point(s) on the graph of

where the slope is 5. (Thanks!)

Hello, Affinity,

to avoid the quotient rule you can first do the division:

$y=\frac{x^2+3x-1}{x}=x+3-\frac{1}{x}$

Now derivate:

$y'=1+\frac{1}{x^2}$

If y' = 5 then you have the equation:

$1+\frac{1}{x^2}=5$ . Solve for x and you'll get the answer topsquark has already calculated.

EB