Determine to point of maximum and minimum of a function

Question

I am learning calculus by watching some videos that I have downloaded some time ago. Some of the videos are missing.I cannot find the part where it teaches point of maximum and minimum. Along with the videos I have some questions. One of the questions is this

determine the local maximum and minimum points of the function $$ g(x) = \frac{1}{3}x^3 - 3x^2 + 8x + 5 $$ Can you say how to solve one so I can solve the others.

Answer 1

Hint: Find derivative and equate it to zero:

$g'(x)=x^2-6x+8=0$

Which gives $x_1=4$ and $x_2=2$

Now put these values in $g(x)$:

$g(4)=..?$

$g(2)=..?$

Answer 2

I'm giving the theory of maximum and minimum first and then you should be able to relate. If you sketch a graph between an independent variable x and a function of that variable f(x) = y (say), and that can be any arbitrary curve. If you carefully investigate the curve, you will find that, at the point on the curve, where it is maximum or minimum, ie. the height of that point above the x-axis is either maximum or minimum, the curve is kind of flat there. More specifically, if you look on the tangent at every point on the curve, you will see that at "such" points, the tangent to the curve is parallel to the x-axis. Conversely, you can say that the point on the curve is maximum or minimum, if the "tangent to the curve at that point" is parallel to the x-axis. To check that mathematically, you could search for the "slope of the tangent on every point on the curve", and slope of tangent to the curve at any point (x, y) on the curve is dy/dx. For tangents parallel to x-axis, it's slope is 0, so dy/dx is 0, and remember y=f(x). Procedure: Evaluate the first derivative of y=f(x), equate that to 0, to find the x for which f'(x) is 0, and then find the corresponding y for that x, and you are done.