The graph of f(x)=-1+(x-1)^2 e^x is shown in Fig. 5. Find the coordinates of the relative maximu

The graph of f(x)=-1+(x-1)^2 e^x is shown in Fig. 5. Find...

Key Concepts

Differentiation Techniques

Differentiation is the process used to compute the derivative of a function, which is essential for analyzing how the function changes. Techniques such as the product rule and chain rule allow one to differentiate complex expressions by breaking them into simpler parts, which is crucial when dealing with functions composed of multiple types of terms.

Critical Points

Critical points of a function are the points in its domain where the derivative is zero or undefined. These points are important because they are potential locations of relative (local) extrema, and analyzing them helps in understanding the function’s overall behavior.

Relative Extrema

Relative extrema refer to the local maximum and minimum values of a function, which are the highest or lowest points in a small region around a point. Identifying these points is key to understanding the local behavior and overall shape of a graph.

First Derivative Test

The first derivative test involves analyzing the sign changes of the derivative around the critical points. A change from positive to negative indicates a relative maximum, while a sign change from negative to positive indicates a relative minimum. This test provides a method for classifying the critical points based on the slope behavior of the function.

Second Derivative Test

The second derivative test uses the value of the second derivative at a critical point to determine the concavity of the function. If the second derivative is positive at a critical point, the function is concave up, confirming a relative minimum; if it is negative, the function is concave down, confirming a relative maximum. This test offers an alternative method for classifying critical points.

Transcript

00:01 In this question we recall that the relative extremer occurs at f -pram on the x equal to 0.

00:17 And here we are given the function f x equal to the minus 1 plus x minus 1 square times e to the x.

00:33 And then here we need to find the first derivative.

00:36 So f prime the x the derivative the minus 1 equal to 0.

00:41 For this one we need to use the product rule.

00:44 So this one will be the u, this will be the v.

00:47 Recorder the u times v prime, it will be equal to u prime v plus u v prime.

00:54 Therefore derivative of x minus 1 square it will equal to the 2 x minus 1 times e to the x then plus we have the x minus 1 square times e e to the x.

01:08 Now if we simplify this one we have e to the x times x minus 1, times with the 2 plus x minus 1.

01:18 So if we simplify we get equal to e to the x, x minus 1, and then we have the x plus 1.

01:28 From here we say this one equal to 0 to find the critical point.

01:32 This implies that the x minus 1 equal to 0 x plus 1 must equal to 0 exponential will never equal to 0 and isn't implies that the x equal to 1 x equal to minus 1 and now if we try to find the critical uh let's try to find the using the first derivative test so we have this will be the x and then we have here will be from us infinity to infinity this will be the f prime the x.

02:12 So we have the value here will be the minus 1 and the 1...

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