question 4\nin the diagram below, the graph of $f(x)=a\\sin bx + 1$ and $g(x)=2\\cos(x + p)$ are shown\nfor…

Explanation:

Step1: Recall amplitude formula for cosine

For $y = A\cos(Bx + C)+D$, amplitude is $|A|$. For $g(x)=2\cos(x + p)$, amplitude $= 2$.

Step2: Recall period formula for sine

For $y = a\sin(bx)+1$, period $T=\frac{360^{\circ}}{|b|}$. From the graph of $f(x)$, the period is $120^{\circ}$, so $\frac{360^{\circ}}{|b|}=120^{\circ}$, then $|b| = 3$. Since the graph is moving as a normal sine - like curve, $b = 3$.

Step3: Find $a$ using a point on $f(x)$

The point $(48.05^{\circ},0.41)$ lies on $f(x)=a\sin(3x)+1$. Substitute $x = 48.05^{\circ}$ and $y = 0.41$: $0.41=a\sin(3\times48.05^{\circ})+1$. $\sin(144.15^{\circ})\approx0.588$. $0.41=a\times0.588 + 1$. $a\times0.588=0.41 - 1=-0.59$. $a=\frac{-0.59}{0.588}\approx - 1$.

Step4: Find $p$ using a point on $g(x)$

The graph of $g(x)=2\cos(x + p)$ passes through $(0,1)$. Substitute $x = 0$ and $y = 1$: $1 = 2\cos(0 + p)$. $\cos(p)=\frac{1}{2}$. In the given domain, $p = 60^{\circ}$.

Step5: Solve $g(x)>0$

$g(x)=2\cos(x + 60^{\circ})>0$. $\cos(x + 60^{\circ})>0$. $-90^{\circ}<x + 60^{\circ}<90^{\circ}$ in one period. $-150^{\circ}<x<30^{\circ}$. In the domain $-90^{\circ}\leq x\leq180^{\circ}$, the solution is $-90^{\circ}\leq x<30^{\circ}$.

Step6: Solve $f(x)\geq g(x)$

From the graph, $f(x)\geq g(x)$ when $-90^{\circ}\leq x\leq - 30^{\circ}$ or $60^{\circ}\leq x\leq120^{\circ}$.

Step7: Solve $f(x)\cdot g(x)<0$

$f(x)\cdot g(x)<0$ when $f(x)$ and $g(x)$ have opposite signs. From the graph, this is when $-30^{\circ}<x<60^{\circ}$ or $120^{\circ}<x\leq180^{\circ}$.

Step8: Analyze $h(x)=h(-x)$

$h(x)=g(x + t)=2\cos((x + t)+60^{\circ})$, $h(-x)=2\cos((-x + t)+60^{\circ})$. Since $h(x)=h(-x)$, the function $h(x)$ is even. For $y = \cos(Ax + B)$, it is even when $B = k\cdot180^{\circ},k\in\mathbb{Z}$. $(x + t)+60^{\circ}=(-x + t)+60^{\circ}+k\cdot360^{\circ}$ or $(x + t)+60^{\circ}=-( - x + t)-60^{\circ}+k\cdot360^{\circ}$. The first equation gives $0 = k\cdot360^{\circ}$ (not useful for finding $t$). The second equation: $x + t+60^{\circ}=x - t - 60^{\circ}+k\cdot360^{\circ}$. $2t=-120^{\circ}+k\cdot360^{\circ}$. $t=-60^{\circ}+k\cdot180^{\circ},k\in\mathbb{Z}$. In the domain (assuming we are looking for values in a reasonable range related to the given domain of $x$), when $k = 0,t=-60^{\circ}$; when $k = 1,t = 120^{\circ}$.

Answer:

4.1. $2$ 4.2. $120^{\circ}$ 4.3. $a=-1,b = 3,p = 60^{\circ}$ 4.4.1. $-90^{\circ}\leq x<30^{\circ}$ 4.4.2. $-90^{\circ}\leq x\leq - 30^{\circ}$ or $60^{\circ}\leq x\leq120^{\circ}$ 4.4.3. $-30^{\circ}<x<60^{\circ}$ or $120^{\circ}<x\leq180^{\circ}$ 4.5. $t=-60^{\circ},120^{\circ}$