Math Problem Statement

solve this

Soit f la fonction définie sur [-3, +∞[ par : f(x) = (2x + 3) / (x + 5). 1) Démontrer que f(x) peut aussi s'écrire : f(x) = 2 - 7 / (x + 5). 2) Démontrer que f est croissante sur [-3, +∞[. 3) a) Démontrer que f admet un minimum, le préciser. b) Démontrer que f admet un majorant, en préciser un. c) En déduire que f est bornée et indiquer un encadrement de f(x).

Solution

Let's go step by step through this exercise:

We are given the function f(x)=2x+3x+5f(x) = \frac{2x + 3}{x + 5}, and the goal is to solve the following problems:

1. Prove that f(x)f(x) can also be written as f(x)=27x+5f(x) = 2 - \frac{7}{x + 5}

We start by simplifying the original function:

f(x)=2x+3x+5f(x) = \frac{2x + 3}{x + 5}

We want to express this in the form f(x)=27x+5f(x) = 2 - \frac{7}{x + 5}. Let's rewrite the fraction:

2x+3x+5=(2x+10)7x+5=2(x+5)7x+5\frac{2x + 3}{x + 5} = \frac{(2x + 10) - 7}{x + 5} = \frac{2(x + 5) - 7}{x + 5}

Now, we can split the fraction:

2(x+5)7x+5=2(x+5)x+57x+5=27x+5\frac{2(x + 5) - 7}{x + 5} = \frac{2(x + 5)}{x + 5} - \frac{7}{x + 5} = 2 - \frac{7}{x + 5}

So, we have shown that:

f(x)=27x+5f(x) = 2 - \frac{7}{x + 5}

2. Prove that f(x)f(x) is increasing on the interval [3,+[[-3, +\infty[

To show that f(x)f(x) is increasing, we need to check the sign of the derivative f(x)f'(x).

Recall f(x)=2x+3x+5f(x) = \frac{2x + 3}{x + 5}. Let's calculate the derivative using the quotient rule:

f(x)=(x+5)(2)(2x+3)(1)(x+5)2f'(x) = \frac{(x + 5)(2) - (2x + 3)(1)}{(x + 5)^2}

Simplifying the numerator:

f(x)=2(x+5)(2x+3)(x+5)2=2x+102x3(x+5)2=7(x+5)2f'(x) = \frac{2(x + 5) - (2x + 3)}{(x + 5)^2} = \frac{2x + 10 - 2x - 3}{(x + 5)^2} = \frac{7}{(x + 5)^2}

Since (x+5)2(x + 5)^2 is always positive for all x>5x > -5, f(x)=7(x+5)2f'(x) = \frac{7}{(x + 5)^2} is positive for x3x \geq -3.

Thus, f(x)f(x) is increasing on the interval [3,+[[-3, +\infty[.

3a. Prove that f(x)f(x) admits a minimum and specify it

Since f(x)f(x) is increasing, the minimum value of f(x)f(x) occurs at the left endpoint of the interval, which is x=3x = -3.

Let's calculate f(3)f(-3):

f(3)=2(3)+33+5=6+32=32=32f(-3) = \frac{2(-3) + 3}{-3 + 5} = \frac{-6 + 3}{2} = \frac{-3}{2} = -\frac{3}{2}

Therefore, the minimum value of f(x)f(x) is f(3)=32f(-3) = -\frac{3}{2}.

3b. Prove that f(x)f(x) admits an upper bound and specify it

Since f(x)f(x) is increasing and unbounded as x+x \to +\infty, there is no finite upper bound. However, for large values of xx, f(x)f(x) approaches 2:

limx+f(x)=2\lim_{x \to +\infty} f(x) = 2

Thus, the function has an asymptote at y=2y = 2, which can be considered its upper bound in the limit.

3c. Deduce that f(x)f(x) is bounded and provide an interval containing f(x)f(x)

From the previous results, we know that f(x)f(x) has a minimum of 32-\frac{3}{2} at x=3x = -3 and approaches 2 as x+x \to +\infty. Therefore, f(x)f(x) is bounded between 32-\frac{3}{2} and 2.

Thus, the interval containing f(x)f(x) is:

f(x)[32,2[f(x) \in \left[-\frac{3}{2}, 2\right[


Would you like further details on any of the steps? Here are some related questions to expand your understanding:

  1. What is the significance of an asymptote in a function like this?
  2. Can you explain how the quotient rule for derivatives works?
  3. How would the behavior of the function change if the denominator were different, say x+3x + 3?
  4. How can you generalize this process for proving whether functions are increasing or decreasing?
  5. Why does the function have a minimum, but not a maximum?

Tip: When working with increasing or decreasing functions, the derivative is a crucial tool. If the derivative is positive, the function is increasing, and if negative, it's decreasing!

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Math Problem Analysis

Mathematical Concepts

Functions
Derivative
Increasing functions
Asymptotes
Boundedness

Formulas

Quotient rule for derivatives
Function transformation

Theorems

Derivative test for increasing functions
Limits and asymptotes

Suitable Grade Level

Grades 11-12