A. Locate the critical points of f b. Use the First...

a. Locate the critical points of $f$
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).
$$f(x)=2 x^{3}+3 x^{2}-12 x+1 ;[-2,4]$$

Key Concepts

Local Extremum

Local extrema refer to local maximum and minimum values of a function, which occur at points where the function reaches a peak or valley compared to surrounding values. Identifying local extrema often involves first finding the critical points and then using tests like the First Derivative Test to classify the type of extremum.

Absolute Extremum on a Closed Interval

The absolute maximum and minimum values of a function on a closed interval are the largest and smallest function values attained over the entire interval, including at the endpoints. To identify these, one must evaluate the function at all critical points within the interval and at the endpoints, then compare these values.

Critical Points

Critical points are values in the domain of a function where its derivative is zero or undefined. They are important because they can indicate where a function changes from increasing to decreasing or vice versa, which may lead to local maximum or minimum values.

First Derivative Test

The First Derivative Test is a technique used to determine whether a critical point corresponds to a local maximum, local minimum, or neither. This involves analyzing the sign changes of the derivative before and after the critical point to assess how the function behaves in its vicinity.

Transcript

00:02 In this question, the function given to us is f of x is equals 2, 2xq plus 3 x square minus 12x plus 1.

00:18 So that is a function.

00:19 And we have to discuss it from minus 2 to 4.

00:23 First you have to calculate the critical point.

00:26 So for critical points, let's differentiate it.

00:28 So f does of x if we calculate it, it will be coming onto it.

00:32 6x square plus 6x minus 12.

00:35 And if you simplify it so it will be 6 into x plus 2 into x minus 1 so that we'll be getting and if we get it to 0 so the critical points are essentially so answer for part 1 we'll be so critical points points are x is equals to minus 2 and x is equals to 1 so these are the two critical points we'll be getting given function now for checking for maximum when you are using first derivative we will see the sign of first derivative okay so we use the sign determination so the critical points are essentially minus 2 and 1 and we can clearly see from here that if we the sign will be positive if you take x greater than 1 and sign will be negative if we take x between minus 2 to 1 and if it will be positive if we take if you take x less than minus 2 to 1 and if it will be positive so, we have to check for the maximum and minima using this sign convention.

01:48 So from here we can clearly see that the sign is from positive to negative at this point.

01:58 Okay, so the graph will be increased and then decrease.

02:01 So at this point we will be getting point of maximum.

02:07 Okay, and here the graph is negative to positive, so it will be decreasing.

02:11 So this will be point of minimum...

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