a. Locate the critical points of ƒ.b. Use the First Derivative Test to locate the local maximum and minimum values.c. Identify the absolute maximum and minimum values of the functionon the given interval (when they exist). ƒ(x) = 2x3 + 3x2 - 12x + 1 on ⌊-2, 4⌋
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.
a. Locate the critical points of ƒ.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Identify the absolute maximum and minimum values of the function
on the given interval (when they exist).
ƒ(x) = 2x3 + 3x2 - 12x + 1 on ⌊-2, 4⌋
Given
ƒ(x) = 2x3 + 3x2 - 12x + 1 on ⌊-2, 4⌋
(a)
Find the critical points first
Put the derivative equal to 0 to find the critical points
(b)
Now with these critical point form the test intervals [-2,1), (1,4]
Apply the first derivative test
First check the sign of the derivative on these test intervals
From the interval [-2,1) choose x = 0 as the test point
and from the interval (1,4) chose x = 2 as the test point
By the first derivative test x = 1 correspond to local minima and the minimum value is -6
Since x=-2 is the boundary point, therefore check the value of function f at x = -2
f(-2) = 21
this implies x = -2 corresponds to local maxima and the maximum value is 21.
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