# Finding derivatives.

• Nov 8th 2010, 09:16 PM
FullBox
Finding derivatives.
Hello everyone. Once again, I've run into problems in my homework that I was unable to solve, and would greatly appreciate some help in figuring them out. I typed them up on Microsoft Word and screen captured them from there.

http://oi55.tinypic.com/vg5qgn.jpg

http://oi56.tinypic.com/2j5b12.jpg

http://oi52.tinypic.com/rvv790.jpg

Thank you in advance to whoever helps out.
• Nov 8th 2010, 10:52 PM
Unknown008
1. Well, I don't know that formula, but this is what I do.

The gradient at any point on a curve is given by the derivative of the curve.

$\displaystyle \dfrac{dy}{dx} = 18x +4$

Now, the gradient is 76. So, the derivative equals 76.

$\displaystyle 76 = 18x + 4$

$\displaystyle x = 4$

Now, plug in the value of x in the equation of the original curve.

2. Convert the upper fraction into one single fraction.

$\displaystyle \dfrac{\frac2x - \frac23}{x- 3} = \dfrac{\frac{6-2x}{3x}}{x-3} = \dfrac{2(3-x)}{3x(x-3)} = -\dfrac{2}{3x}$

Try out part b now.

3. To get the x coordinates of the critical poinst, set the derivative (f'(x) not f(x) as in your post. I think this was a typo) to 0 and solve for x.

Then, if it is like the cube function, is there a maximum? A minimum perhaps? Or neither?
• Nov 8th 2010, 11:19 PM
FullBox
Quote:

Originally Posted by Unknown008
1. Well, I don't know that formula, but this is what I do.

The gradient at any point on a curve is given by the derivative of the curve.

$\displaystyle \dfrac{dy}{dx} = 18x +4$

Now, the gradient is 76. So, the derivative equals 76.

$\displaystyle 76 = 18x + 4$

$\displaystyle x = 4$

Now, plug in the value of x in the equation of the original curve.

9(x)^2 + 4(x) -> 9(4)^2 + 4(4) = 9(16) + 16 = 160

Thanks a ton man. I didn't know that all I had to do was get the derivative of the function, and make it equal to 76 to solve for x. I was WAY off the way I was trying to do it before.

Quote:

2. Convert the upper fraction into one single fraction.

$\displaystyle \dfrac{\frac2x - \frac23}{x- 3} = \dfrac{\frac{6-2x}{3x}}{x-3} = \dfrac{2(3-x)}{3x(x-3)} = -\dfrac{2}{3x}$

Try out part b now.
y - (2/3) = -(2/9)(x-3) -> y = -(2/9)x + 6/9 + 2/3 -> y = -2/9x + 4/3

Quote:

3. To get the x coordinates of the critical poinst, set the derivative (f'(x) not f(x) as in your post. I think this was a typo) to 0 and solve for x.

Then, if it is like the cube function, is there a maximum? A minimum perhaps? Or neither?

Yeah I just noticed when you pointed it out, I did indeed make a typo. It was supposed to be f'(x).

And the cube function doesn't have a minimum nor a maximum, so then it would be safe to say that there are no critical points on it.
• Nov 8th 2010, 11:38 PM
Unknown008
Yes, for the first 2.

For 3, there is a critical point, it's only that it's neither a maximum nor a minium but is called an inflection point.