The exercise in the image asks to determine the "ensemble de définition" (domain of definition) for several mathematical functions. Here's how we can handle this step-by-step.

The "ensemble de définition" refers to the set of all values of $x$ for which the function is well-defined (i.e., does not involve division by zero, does not take square roots of negative numbers, etc.).

Let's go through the functions one by one:

a) $f(x) = \sqrt{x^3 + x^2 - x}$

• The square root function is only defined for non-negative values inside the square root.
• Therefore, $x^3 + x^2 - x \geq 0$.
• We need to solve this inequality to find the domain.

b) $f(x) = \sqrt{-x^3 - x^2 + 17x - 15}$

• Similarly, the square root requires the expression inside to be non-negative.
• Thus, $-x^3 - x^2 + 17x - 15 \geq 0$.
• Solving this inequality will give us the domain.

c) $f(x) = \sqrt{x^2 - |x - 20|}$

• The square root requires $x^2 - |x - 20| \geq 0$.
• Since it involves an absolute value, this needs to be handled carefully.
• We'll need to break the expression into cases depending on the sign of $x - 20$.

d) $f(x) = \sqrt{\frac{x^2 - x - 2}{x^2 - x + 1}}$

• The square root must have a non-negative argument: $\frac{x^2 - x - 2}{x^2 - x + 1} \geq 0$.
• Also, $x^2 - x + 1 \neq 0$ because division by zero is undefined.
• Solve this inequality to find the domain.

e) $f(x) = \sqrt{\frac{1 + x}{x^3 - 1}}$

• The square root function requires $\frac{1 + x}{x^3 - 1} \geq 0$.
• Also, $x^3 - 1 \neq 0$, i.e., $x \neq 1$, to avoid division by zero.
• Solve for the values of $x$.

f) $f(x) = \frac{[E(x)]^2 - 3E(x) + 2}{x^2 + 1}$

• The denominator $x^2 + 1 \neq 0$ for all real $x$ (since it's always positive), so no division by zero here.
• The domain is all real numbers unless there's a restriction from $E(x)$ (which is not clear from the image).

g) $f(x) = \sqrt{x - E(x)}$

• The square root requires $x - E(x) \geq 0$, which depends on $E(x)$ (possibly an external function not fully visible here).
• Solve the inequality to find the domain.

h) $f(x) = \frac{\sqrt{1 - x} - \sqrt{1 + x}}{\sin(x)}$

• The square roots require $1 - x \geq 0$ and $1 + x \geq 0$, meaning $-1 \leq x \leq 1$.
• Additionally, $\sin(x) \neq 0$, so exclude points where $\sin(x) = 0$ (i.e., $x = 0, \pi, 2\pi, \ldots$).

i) $f(x) = \frac{\sqrt{x^2 - x + 6}}{|3x - 1| - 4}$

• The square root requires $x^2 - x + 6 \geq 0$, which is always true because this quadratic expression is always positive.
• However, $|3x - 1| - 4 \neq 0$, so solve for values where $|3x - 1| \neq 4$.

j) $f(x) = x - \sqrt{\frac{E(x)}{2}}$

• The square root requires $\frac{E(x)}{2} \geq 0$, so $E(x) \geq 0$.
• Solve for $x$ based on this condition.

Would you like detailed solutions to each specific part or help solving a particular inequality?

5 Related Questions:

1. How do we solve inequalities involving square roots?
2. What happens when the argument inside a square root is negative?
3. How can we handle absolute value expressions within functions?
4. What is the method to solve rational inequalities like the one in part $d$?
5. How do we determine the domain when trigonometric functions are involved?

Tip:

Always check for division by zero and negative square roots when determining the domain of functions!