Find the critical points for each function. Use the first derivative test to determine whether t

Find the critical points for each function. Use the first...

Key Concepts

First Derivative Test

The first derivative test is a method used to classify critical points. By analyzing the sign change of the derivative before and after a critical point, one can determine whether the point is a local maximum (derivative changes from positive to negative), a local minimum (derivative changes from negative to positive), or neither (no sign change).

Differentiation Techniques

Differentiation techniques such as the power rule, product rule, quotient rule, and chain rule are essential to find the derivative of a function. Mastery of these techniques allows for the computation of derivatives of a wide variety of functions, enabling the identification of critical points and analysis of function behavior.

Derivative

The derivative of a function measures its rate of change and is fundamental in calculus for analyzing the behavior of functions. It provides information about the slope of the function at any point, which is crucial for identifying turning points, increasing or decreasing intervals, and for solving optimization problems.

Critical Points

Critical points are the values in the domain of a function where its derivative is zero or undefined. These points are important because they are candidates for local maxima, local minima, or points of inflection, and they often indicate where the function's behavior changes.

Transcript

00:01 For this problem, we want to find the critical points for each function, which will write out as we go through.

00:07 And we want to use the first derivative test to determine the nature of the critical point.

00:12 So, our first function is y equals x to the power of 4 minus 8x squared.

00:19 Now, to find the critical points, we take the derivatives.

00:22 So that's going to be 4x cubed minus 16x.

00:27 And then to actually find the critical points, we set that equal to zero.

00:30 First thing we can see is that we have a common factor of 4x here between those two terms, so we can factor that out.

00:37 So we get 4x times x squared minus 4.

00:42 And that's going to equal 0, which tells us we'll have a critical point when x equals 0.

00:48 And then we'll also have critical points when x squared minus 4 equals 0, or x squared equals 4, or x equals plus or minus 2, taking the square root of both.

01:01 Size there.

01:03 So we have critical points at negative 2, 0, and 2.

01:08 Now, to apply the first derivative test, since we have this positive coefficient on the leading term here, a positive 1, we'll be coming in from positive values, which means that between negative 2 and 0, we'll go from positive to negative, then we'll go back to positive, then we'll go to negative.

01:30 Or excuse me, one moment.

01:34 Excuse me, need to be careful here.

01:37 We have this positive coefficient on the leading term for the first derivative, this 4x cubed.

01:43 But since we are coming in from x cubed, that does mean that we'll actually be coming in from negative values, because we will be coming in from large negative values and a negative cube will still be a negative.

01:56 So we end up going negative, positive, negative, positive.

02:00 Which means if we go from having a negative first derivative to a positive first derivative across this point negative 2, then that means that we are going to have a minimum.

02:16 Then 0 is going to be a relative max or a local max, say? and then at 2 we'll have another minimum.

02:30 Now for part b, we have f of x equals 2x...

Please give Ace some feedback