Thread: Show that dy/dx is +ve when a < x < b

1. Show that dy/dx is +ve when a < x < b

I have a differentiation problem which starts with

$y = x^2 (4 - x)$

I have to find dy/dx. No problem there, get

$f'(x) = 8x - 3x^2$

Then to find values of x where dy/dy = 0

$x(8 - 3x) = 0$
$x = 0, x = \frac{8}{3}$

But then it asks "Denoting the values of x which you have just calculated by a and b, where a < b, show dy/dx is positive when a < x < b".

I can see graphically where dy/dx is positive, but do I need to show algebraically? How would I answer this?

Angus

2. Re: Show that dy/dx is +ve when a < x < b

Originally Posted by angypangy
I have a differentiation problem which starts with

$y = x^2 (4 - x)$

I have to find dy/dx. No problem there, get

$f'(x) = 8x - 3x^2$

Then to find values of x where dy/dy = 0

$x(8 - 3x) = 0$
$x = 0, x = \frac{8}{3}$

But then it asks "Denoting the values of x which you have just calculated by a and b, where a < b, show dy/dx is positive when a < x < b".

I can see graphically where dy/dx is positive, but do I need to show algebraically? How would I answer this?

Angus
You want to find where \displaystyle \begin{align*} 8x - 3x^2 > 0 \end{align*}

\displaystyle \begin{align*} 8x - 3x^2 &> 0 \\ 3x^2 - 8x &< 0 \\ 3\left(x^2 - \frac{8}{3}x\right) &< 0 \\ 3\left[x^2 - \frac{8}{3}x + \left(-\frac{4}{3}\right)^2 - \left(-\frac{4}{3}\right)^2\right] &< 0 \\ 3\left[\left(x - \frac{4}{3}\right)^2 - \frac{16}{9}\right] &< 0 \\ \left(x - \frac{4}{3}\right)^2 - \frac{16}{9} &< 0 \\ \left(x - \frac{4}{3}\right)^2 &< \frac{16}{9} \\ \left|x - \frac{4}{3}\right| &< \frac{4}{3} \\ -\frac{4}{3} < x - \frac{4}{3} &< \frac{4}{3} \\ 0 < x &< \frac{8}{3} \end{align*}

So the derivative is positive in between these values.