QuestionSoit g la fonction définie par $g(x)=\frac{\sqrt{x-3}-2}{\sqrt{x-3}+2}$ $\left.1^{\circ}\right)$ Donner le domaine de définition de $g$ . $2^{\circ}$ ) a) Calculer $g(x)-g(3)$ b) Déduire que $g$ admet un minimum en 3 $\left.3^{\circ}\right)$ Montrer que $g$ est majoré par 1

Studdy Solution

STEP 1

1. The function $g(x) = \frac{\sqrt{x-3} - 2}{\sqrt{x-3} + 2}$ involves square roots and a fraction.
2. The domain of $g(x)$ is determined by the values of $x$ for which the expression is defined, specifically where the square root is non-negative and the denominator is not zero.
3. To find $g(x) - g(3)$ , we will substitute and simplify.
4. To show that $g$ has a minimum at 3, we will use the result from part 2a.
5. To show that $g$ is bounded above by 1, we will analyze the expression.

STEP 2

1. Determine the domain of $g(x)$ .
2. Calculate $g(x) - g(3)$ .
3. Deduce that $g$ has a minimum at 3.
4. Show that $g$ is bounded above by 1.

STEP 3

Determine the domain of $g(x)$ :
The expression $\sqrt{x-3}$ is defined for $x - 3 \geq 0$ , which means:
$x \geq 3$
Additionally, the denominator $\sqrt{x-3} + 2$ must not be zero. Since $\sqrt{x-3} \geq 0$ , $\sqrt{x-3} + 2$ is never zero for $x \geq 3$ .
Thus, the domain of $g(x)$ is:
$x \geq 3$

STEP 4

Calculate $g(x) - g(3)$ :
First, find $g(3)$ :
$g(3) = \frac{\sqrt{3-3} - 2}{\sqrt{3-3} + 2} = \frac{0 - 2}{0 + 2} = \frac{-2}{2} = -1$
Now, calculate $g(x) - g(3)$ :
$g(x) - g(3) = \frac{\sqrt{x-3} - 2}{\sqrt{x-3} + 2} + 1$
Simplify the expression:
$= \frac{\sqrt{x-3} - 2 + (\sqrt{x-3} + 2)}{\sqrt{x-3} + 2}$ $= \frac{2\sqrt{x-3}}{\sqrt{x-3} + 2}$

STEP 5

Deduce that $g$ has a minimum at 3:
From the expression $g(x) - g(3) = \frac{2\sqrt{x-3}}{\sqrt{x-3} + 2}$ , note that for $x = 3$ , the expression is zero. For $x > 3$ , $\sqrt{x-3} > 0$ , making the expression positive. Thus, $g(x) \geq g(3)$ for $x \geq 3$ .
Therefore, $g$ has a minimum at $x = 3$ .

STEP 6

Show that $g$ is bounded above by 1:
Consider the expression $g(x) = \frac{\sqrt{x-3} - 2}{\sqrt{x-3} + 2}$ .
For $x \geq 3$ , $\sqrt{x-3} \geq 0$ , so:
$\sqrt{x-3} - 2 \leq \sqrt{x-3} + 2$
Thus:
$\frac{\sqrt{x-3} - 2}{\sqrt{x-3} + 2} \leq 1$
Therefore, $g(x) \leq 1$ for all $x \geq 3$ .
The domain of $g$ is $x \geq 3$ , $g$ has a minimum at $x = 3$ , and $g$ is bounded above by 1.

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