# Function derivatives

Earlier I asked a question about derivatives and for some reason i'm just not able to answer the question. However, I've attempted a near identical question but on another past paper, and think I may have it. If someone could check over and see if it's correct. If it is correct, could someone explain to me why I can do this one but not the other -.-

Given the function $f(x)=2x^3-3x^2-36x+5$

1. Compute the derivative $f'(x)$
2. Find and classify the stationary points of $f(x)$

$$\frac{d}{dx} f(x) = 6x^2 - 6x - 36$$

I then used the quadratic equation in order to find the values of $x$, being $-2$ when minus and $3$ when addition. So, I then went onto to calculate $y$ for each by substituting the values of $x$ in. From this I learned when $x = -2, y = 49$ and when $x = 3, y = -76$. Giving me $( -2, 49 )$ and $( 3, -76 )$.

From here I introduced derivative 2 in order to find rate of change.

From this I found $$\frac{d^2y}{dx^2} = 12x - 6$$

Once again I substituted the values of $x$ in and found when $x = -2 dx^2 = -30$ meaning it is in fact a maximum since it's below the value 0.

I then went onto to prove $x = 3$ eventually giving me that $dx^2$ is $30$ therefore $(3, -76)$ is a minimum.

I'm pretty shit at maths to be honest. However, I found this pretty straight forward. However, for my other question Quadratic formula - math error I still feel literally clueless. If I understand this then should I understand my other question? The questions are basically identical, I just don't understand the logic..

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+1 for curiosity and persistence. You need to look at second derivative. – user114628 Jan 9 '14 at 1:11
persistence is required when it's an exam topic, dude :D otherwise i'd bed in bed now hehe. Also, are you referring to the current thread or my previous one ? i'm unsure whether you're implying my second derivative is wrong in my post? – user119325 Jan 9 '14 at 1:15
What are the parts of the method I briefly explained in Quadratic formula - math error that you still don't feel/think you understand? – Dror Jan 9 '14 at 1:16
@user2943324 meant you need to study about the 2nd derivative, what it means, how does it affect the function, why is it useful, when is it useful, what is it intuitively.. etc.. – Dror Jan 9 '14 at 1:19
My first problem is when using quadratic formulas i'm given two values for X - since you two do one minus and another addition. 2X-2=0 meaning x = 1 , no quadratic formula. However we only have a single x rather than 2 ? I'm pretty confused by that – user119325 Jan 9 '14 at 1:25

$$x^2 - 2 x + 4$$ So that the first derivative were: $$2x - 2 = 0$$ and solving this $x=1$ the stationary point were: $$(1,1^2-2\cdot1+4=3)$$ and calculating the second derivative you will obtain: $$2 > 0$$ which say you: the stationary point is a minimum. The logic is the same but the function are different, and calculations go in different ways.
Here you can see the graph for the two function and their derivative. In the old situation you have just one minimum, in fact the line for hte first derivative have just one intersection with x axis in the point $x=1$. In this new situation you have a cubic which derivative is a quadratic function with two intersection with x axis in $x=3$ and $x=-2$. And that's all.