Let's go through each problem from the image you provided:

1. Tentukan penyelesaian pertidaksamaan $\frac{2x + 1}{x - 3} > 1$

To solve the inequality:

$\frac{2x + 1}{x - 3} > 1$

• Subtract 1 from both sides: $\frac{2x + 1}{x - 3} - 1 > 0$

• Combine the fractions on the left-hand side: $\frac{2x + 1 - (x - 3)}{x - 3} > 0$ $\frac{x + 4}{x - 3} > 0$

• The critical points are $x = -4$ and $x = 3$.

• Test intervals around these critical points to determine where the inequality holds.

2. Tentukan domain dari fungsi $f(x) = \sqrt{\frac{2x + 1}{x^2 - 3}}$

For the domain of this function:

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

• The expression inside the square root must be non-negative: $\frac{2x + 1}{x^2 - 3} \geq 0$

• Also, the denominator $x^2 - 3$ cannot be zero, so solve for $x^2 - 3 = 0$, which gives the restriction $x \neq \pm\sqrt{3}$.

• Analyze the sign of the expression to find the intervals where the function is defined.

3. Nyatakan fungsi $h(x) = x \sin(3x + 4)$ berikut sebagai komposisi dua fungsi $(f \circ g)(x)$ dalam tiga cara

Here, $h(x) = x \sin(3x + 4)$ can be expressed as the composition of functions in several ways. For example:

1. $g(x) = 3x + 4$ and $f(u) = x \sin(u)$, so $h(x) = (f \circ g)(x)$.
2. $g(x) = x$ and $f(u) = u \sin(3u + 4)$.
3. $g(x) = x$ and $f(u) = x \sin(3x + u)$.

4. Gambarlah grafik fungsi $f(x) = (x - 1)^2 - 2$

This is a quadratic function:

$f(x) = (x - 1)^2 - 2$

• The graph is a parabola that opens upwards.
• The vertex is at $(1, -2)$.

Would you like the solutions to be worked out in further detail or any additional help with a specific problem?

Relative Questions:

1. How do we determine the intervals for which the inequality holds in question 1?
2. What methods can be used to test the sign of the function in question 2?
3. Can the composition of functions in question 3 be interpreted differently in another context?
4. How does the transformation affect the graph of the quadratic function in question 4?
5. What are some alternative methods for solving the inequality in question 1?

Tip: When solving inequalities, always be careful with the direction of the inequality sign when multiplying or dividing by negative numbers.