To find extrema we need find where the first derivative is equal to zero.
[tex]f'(x)=(3x^3-36x-7)'=9x^2-36[/tex]
We now have to find zeros of this equation:
[tex]9x^2-36=0\\ x^2=\frac{36}{9}\\ x^2=4\\ x_1=2\\ x_2=-2[/tex]
These are x coordinates of our extrema, we calculate y coordinates simply by plugging in x coordinates in our functions:
[tex]f(2)=3(2)^3-36(2)-7=-55\\ f(-2)=3\left(-2\right)^3-36\left(-2\right)-7=41[/tex]
Function incrases when it's first derivative is positive.
Our extrema points are (-2,41) and (2,-55).
This is because the first derivative gives you the slope of the function.
We need to find where our first derivative is positive.
The first derivative is parabola that has positive coefficient a. Parabolas with positive coefficient a ("happy" parabolas or concave) are negative between zeros and positive everywhere else. So our function is increasing when x<-2 and x>2.
In order to determine where the function is concave up or down, we have to find the second derivative.
[tex]f''(x)=(f'(x))'=(9x^2-36)'=18x[/tex]
When the second derivative is positive function is concave up, whet it is negative function is concave down.
Our function is concave up when x>0 and it is concave down when x<0.
