Math Problem Statement

La courbe ci-contre est celle d'une fonction f numérique : 1) Déterminer D l'ensemble de définition de f. 2) f est-elle continue sur D? 3) Déterminer f([-3; -1]), f(] -3; -1[), f([-1; 5]), f([-3; 3]) et f([-3; +∞[).

Solution

Let's analyze and solve this exercise based on the image provided:

The problem is based on a function ff whose graph is given. Let's go through each question in detail.

1) Determine DD, the domain of definition of ff.

From observing the graph, it appears that the function ff is defined from x=3x = -3 to x=5x = 5, inclusive. Thus, the domain DD of ff is: D=[3,5]D = [-3, 5]

2) Is ff continuous on DD?

To determine continuity, we need to check if there are any breaks, jumps, or holes in the graph of ff over its domain DD. Based on the provided graph, ff appears to be a smooth, continuous curve from x=3x = -3 to x=5x = 5, with no interruptions. Therefore, ff is continuous on DD.

3) Determine the values of ff on various intervals:

Let’s evaluate the ranges of ff over the specified intervals:

  • f([3,1])f([-3, -1]):
    Observing the graph from x=3x = -3 to x=1x = -1, the function starts at f(3)2f(-3) \approx 2 and increases to a maximum value at x=1x = -1 (approximately 4).
    So, f([3,1])=[2,4]f([-3, -1]) = [2, 4].

  • f(3,1)f(-3, -1):
    This is the open interval, so it excludes the endpoints f(3)f(-3) and f(1)f(-1). The range within this interval is similar, but strictly between 2 and 4.
    Thus, f((3,1))=(2,4)f((-3, -1)) = (2, 4).

  • f([1,5])f([-1, 5]):
    From x=1x = -1 to x=5x = 5, the function decreases to a minimum at around y=1y = -1 (near x=2x = 2) and then increases again to around f(5)3f(5) \approx 3.
    Hence, f([1,5])=[1,4]f([-1, 5]) = [-1, 4].

  • f([3,3])f([-3, 3]):
    For this interval, the function ranges from f(3)2f(-3) \approx 2 to a minimum near y=1y = -1 around x=2x = 2.
    Thus, f([3,3])=[1,4]f([-3, 3]) = [-1, 4].

  • f([3,+))f([-3, +\infty)):
    Since the function is only defined up to x=5x = 5, we interpret f([3,+))f([-3, +\infty)) as f([3,5])f([-3, 5]), which spans the full range of the function. The minimum value is around 1-1 and the maximum around 44.
    Therefore, f([3,+))=[1,4]f([-3, +\infty)) = [-1, 4].

Would you like further explanations on any part of this solution?

Here are five follow-up questions to deepen understanding:

  1. How would the domain change if the function graph extended beyond x=5x = 5?
  2. What if there were discontinuities in the graph? How would that affect the continuity on DD?
  3. Can we determine the exact values at specific points (like f(2)f(2)) from this graph?
  4. What kind of function (e.g., polynomial, exponential) does this graph resemble, based on its shape?
  5. How would you determine the derivative of ff if you only had the graph?

Tip: When analyzing function ranges over intervals, carefully check the graph for local minima and maxima within each interval.

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Math Problem Analysis

Mathematical Concepts

Functions
Continuity
Domain and Range

Formulas

-

Theorems

Definition of function continuity
Domain and range of a function

Suitable Grade Level

High School (Grades 10-12)