Let f(x) = 3x^4 − 2x^3. Find all the critical points and determine if they are relative maximums, relative minimums, or neither. Then find the absolute max and absolute min on the interval [−1, 1]. Show and explain all your work.
Minimization
In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in terms of global or local range. The definition of minimization in the thesaurus is the process of reducing something to a small amount, value, or position. Minimization (noun) is an instance of belittling or disparagement.
Maxima and Minima
The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.
Derivatives
A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which one would be easier? Walking on the road would be much easier than climbing a mountain.
Concavity
In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the first derivative test and second derivative test to understand the concave behavior of the function.
Let f(x) = 3x^4 − 2x^3. Find all the critical points and determine if they are
First find the critical point.
Obtain the first derivative of the given function as follows.
Equate the first derivative to zero and solve for x.
Therefore, the critical points are .
These points divide the interval [-1,1] into three intervals .
Choose test points from each interval and substitute those points in the first derivative to find the sign.
| Interval | |||
| Test point | |||
| Value of the first derivative | -3 | ||
| Sign | Negative | Negative | Positive |
Note that, at x=0 there is no sign change and at x= the sign of the derivative changes from negative to positive.
Therefore, there is a relative minimum at x= and a saddle point at x=0.
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