Find the point(s) on the graph of where the slope is 5. (Thanks!)
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: .$\displaystyle \left(\frac{1}{256},\,\frac{13}{256}\right)$
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
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