Math Problem Statement

Determinați care din punctele 0, -1, 1, -1/2, 1/2 sunt puncte de extrem local ale funcției f: R → R, f(x) = x^2 + 2x. Aflati extremele locale respective ale funcției.

Solution

To solve this problem, let’s go through each part step by step.

Problem Analysis

Given:

  1. A function f:RRf : \mathbb{R} \rightarrow \mathbb{R} defined by: f(x)=x2+2xf(x) = x^2 + 2x
  2. Points to examine for extremal (maximum or minimum) behavior: 0,1,1,12,120, -1, 1, -\frac{1}{2}, \frac{1}{2}.

Part (a): Determine Which Points Are Local Extrema

To determine which of these points are local extrema, we need to:

  1. Find the derivative of f(x)f(x) and set it to zero to find the critical points.
  2. Evaluate the second derivative at each critical point or use the first derivative test to classify them as local minima, maxima, or neither.

Step 1: Find the Derivative of f(x)f(x)

f(x)=x2+2xf(x) = x^2 + 2x The first derivative is: f(x)=2x+2f'(x) = 2x + 2 Setting f(x)=0f'(x) = 0 to find the critical points: 2x+2=02x + 2 = 0 x=1x = -1 Thus, x=1x = -1 is a critical point of f(x)f(x).

Step 2: Evaluate the Second Derivative

The second derivative of f(x)f(x) is: f(x)=2f''(x) = 2 Since f(x)=2>0f''(x) = 2 > 0 for all xx, the function is concave up everywhere, indicating that x=1x = -1 is a local minimum.

Conclusion for Part (a)

Among the points 0,1,1,12,120, -1, 1, -\frac{1}{2}, \frac{1}{2}, only x=1x = -1 is a local extremum (specifically, a local minimum).

Part (b): Find the Local Extremum Value

Since we have identified x=1x = -1 as a local minimum, we evaluate f(x)f(x) at x=1x = -1 to find the extremal value:

f(1)=(1)2+2(1)=12=1f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1

Summary

  • Part (a): The point x=1x = -1 is a local extremum.
  • Part (b): The local minimum value at x=1x = -1 is 1-1.

Would you like further details on the solution steps?

Here are 5 related questions for deeper understanding:

  1. How would you find local extrema for more complex functions?
  2. What changes if f(x)f(x) had a different leading coefficient, e.g., f(x)=ax2+bx+cf(x) = ax^2 + bx + c?
  3. How does the second derivative test help determine concavity?
  4. What if we had to find global extrema on a closed interval?
  5. How can the first derivative test be used as an alternative to the second derivative test?

Tip: Always check the concavity using the second derivative, especially if you need to classify the nature of critical points accurately.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Derivative
Critical Points
Second Derivative Test

Formulas

f(x) = x^2 + 2x
First derivative: f'(x) = 2x + 2
Setting f'(x) = 0 to find critical points
Second derivative: f''(x) = 2

Theorems

Second Derivative Test

Suitable Grade Level

Grades 10-12