In the diagram, f(x) = 2 (x-45^) is drawn for the interval x [-180^ ; 180^]. Write down the amplitude of f.

Given $f(x) = 2 \sin(x - 45^\circ)$.
The derivative is $f'(x) = \frac{d}{dx} [2 \sin(x - 45^\circ)] = 2 \cos(x - 45^\circ)$.

We need to solve $2 \cos(x - 45^\circ) \ge 0$, which simplifies to $\cos(x - 45^\circ) \ge 0$.

Let $\theta = x - 45^\circ$. The cosine function is non-negative in the first and fourth quadrants.
So, $-90^\circ + 360^\circ n \le \theta \le 90^\circ + 360^\circ n$, where $n$ is an integer.

Add $45^\circ$ to all parts:

For $n=0$:

This interval is within $[-180^\circ; 180^\circ]$.
For other integer values of $n$, the intervals will fall outside $[-180^\circ; 180^\circ]$.

• At $x=-180^\circ$, $f(-180^\circ) = \sqrt{2}$ and $g(-180^\circ) = 0$. So $f > g$.
• As $x \to -90^\circ$, $f(x) \to -\sqrt{2}$ and $g(x) \to -\infty$. So $f > g$.
• In this interval, $f(x)$ decreases from $\sqrt{2}$ to $-\sqrt{2}$, while $g(x)$ decreases from 0 to $-\infty$. $f(x)$ remains above $g(x)$. No intersections.

• As $x \to -90^\circ$, $f(x) \to -\sqrt{2}$ and $g(x) \to \infty$. So $g > f$.
• At $x=0^\circ$, $f(0^\circ) = -\sqrt{2}$ and $g(0^\circ) = 0$. So $g > f$.
• At $x=45^\circ$, $f(45^\circ) = 0$ and $g(45^\circ) = -1$. So $f > g$.
• Since $g > f$ at $0^\circ$ and $f > g$ at $45^\circ$, there must be one intersection between $0^\circ$ and $45^\circ$.
• As $x \to 90^\circ$, $f(x) \to \sqrt{2}$ and $g(x) \to -\infty$. So $f > g$.
• In $(-90^\circ, 0^\circ)$, $g(x)$ is positive and $f(x)$ is negative. No intersections.
• In $(45^\circ, 90^\circ)$, $g(x)$ is negative and $f(x)$ is positive. No intersections.

• As $x \to 90^\circ$, $f(x) \to \sqrt{2}$ and $g(x) \to \infty$. So $g > f$.
• At $x=135^\circ$, $f(135^\circ) = 2$ and $g(135^\circ) = 1$. So $f > g$.
• Since $g > f$ as $x \to 90^\circ$ and $f > g$ at $135^\circ$, there must be one intersection between $90^\circ$ and $135^\circ$.
• At $x=180^\circ$, $f(180^\circ) = \sqrt{2}$ and $g(180^\circ) = 0$. So $f > g$.
• In $(135^\circ, 180^\circ]$, $f(x)$ decreases from 2 to $\sqrt{2}$, while $g(x)$ decreases from 1 to 0. Since $f(x)$ is always above $g(x)$ at $135^\circ$ and $180^\circ$, and both are decreasing, $f(x)$ remains above $g(x)$. No intersections.

Total number of solutions = $0 + 1 + 1 = 2$.

The graph $g(x) = -\tan x$ is translated $45^\circ$ to the right. This means replacing $x$ with $(x - 45^\circ)$.
So, $h(x) = -\tan(x - 45^\circ)$.

The asymptotes of $\tan \theta$ are at $\theta = 90^\circ + 180^\circ n$.
For $h(x)$, we set $x - 45^\circ = 90^\circ + 180^\circ n$.

• For $n=0$: $x = 135^\circ$. This is in the interval.
• For $n=-1$: $x = 135^\circ - 180^\circ = -45^\circ$. This is in the interval.
• For $n=1$: $x = 135^\circ + 180^\circ = 315^\circ$. This is outside the interval.
• For $n=-2$: $x = 135^\circ - 360^\circ = -225^\circ$. This is outside the interval. The equations of the asymptotes of $h$ in the given interval are $x = -45^\circ$ and $x = 135^\circ$.