Function derivatives

Question

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 bad 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..

Answer 1

In the other excercise the function were:

$$ x^2 - 2 x + 4 $$ 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.