1. ## differentiation problem

Given that $y= \frac{1}{3}x^3-9x$

This part asks me to differentiate the equation and find the minimum and maximums, but I did all that, byt the next part got me:

Given that $24x+3y+2=0$ is the equation of the tangent to the curve at the point (p,q) , find p and q.

Now I got the answer to be (1,-22/3) and (-1, 26/3)

But theres only one answer which is (1,-22/3) and the other one isn't mentioned in the answer book. I know that the tangent is only hitting the curve at one point, but what can help me decide between the two? (by the way I used the method where I found what values of x made the derivative = -8, rather than solving the 2 simultaneously because Im bad at factorising cubics.)

2. Originally Posted by david18
Given that $\displaystyle y= \frac{1}{3}x^{3}-9x$

This part asks me to differentiate the equation and find the minimum and maximums, but I did all that, byt the next part got me:

Given that $\displaystyle 24x+3y+2=0$ is the equation of the tangent to the curve at the point (p,q) , find p and q.

Now I got the answer to be (1,-22/3) and (-1, 26/3)

But theres only one answer which is (1,-22/3) and the other one isn't mentioned in the answer book. I know that the tangent is only hitting the curve at one point, but what can help me decide between the two? (by the way I used the method where I found what values of x made the derivative = -8, rather than solving the 2 simultaneously because Im bad at factorising cubics.)

The tags are 'math', not 'latex'.

The given line equation, when solved for y, yields $\displaystyle y=-8x-\frac{2}{3}$

$\displaystyle \frac{1}{3}x^{3}-9x = -8x-\frac{2}{3}$

Solving this gives us x=1 or x=-2.

But one of these may just be an intersection point. The slope of said line is -8.

So, where does the curve have slope -8?.

$\displaystyle x^{2}-9=-8$

x=1

3. Originally Posted by david18
Now I got the answer to be (1,-22/3) and (-1, 26/3)

I get $\displaystyle (1; - \frac{26}{3} ) \ and \ (-1 ; \frac{22}{3})$

Secondly, if $\displaystyle x^2 = 1$, you are not allowed to say $\displaystyle x = \sqrt{1} = \pm 1$ (That's what i was taught anyway).

4. Originally Posted by galactus
$\displaystyle \frac{1}{3}x^{3}-9x = -8x-\frac{2}{3}$
Galactus shouldn't we be using $\displaystyle x^2 - 9$ ?

The tangent to the derivative of the original equation? Isn't that what he's asking?

5. Originally Posted by janvdl
Galactus shouldn't we be using $\displaystyle x^2 - 9$ ?

The tangent to the derivative of the original equation? Isn't that what he's asking?
Absolutely, janvdl. That's why I amended my post.