1. Calculus Problem

Find the point(s) on the graph of where the slope is 5. (Thanks!)

2. Hello, Affinity!

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

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

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

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

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

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

3. helo everybody
siriban u have written a different function i think

4. sorry thats soroban not siriban!!!

5. Originally Posted by Affinity
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.

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

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

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

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

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

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

$\displaystyle 4x^2 - 1 = 0$

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

Thus $\displaystyle x = \pm 1/2$

-Dan

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

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

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

8. Originally Posted by Affinity
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:

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

Now derivate:

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

If y' = 5 then you have the equation:

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

EB