smc-6217-60-Other problems

Functions, 2ADV EQ-Bank 14

The function $y=f(x)$ is defined by:

\begin{align*}
f(x)= \begin{cases}|x+1|-2, & \text { for }\ x \leqslant 1 \\ x^2-4, & \text { for }\ x>1\end{cases}
\end{align*}

Sketch $y=f (x)$ (3 marks)

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For what values of $x$ is $f(x)=0$? (1 mark)

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a.

b. $f(x)=0\ \ \text{when}\ \ x=-3,2$.

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a.

b. $f(x)=0\ \ \text{when}\ \ x=-3,2$.

Functions, 2ADV EQ-Bank 28

Consider the function

\begin{align*}
f(x)=\begin{cases}2^x, & \text {for }\ x<0 \\ m x+c, & \text {for }\ 0 \leq x \leq 2 \\ \dfrac{8}{x}, & \text {for }\ x>2\end{cases}
\end{align*}

Given that $f(x)$ is continuous at both $x =0$ and $x =2$:

Find the values of $m$ and $c$. (2 marks)

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Identify any asymptotes of $y=f(x)$. (1 mark)

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a. $f(x)=\left\{\begin{array}{cl}2^x & \text {for } x<0 \\ m x+c & \text {for } 0 \leq x \leq 2 \\ \dfrac{8}{x} & \text {for } x>2\end{array}\right.$

$\text {Continuous at}\ \ x=0:$

$2^\circ=m(0)+c \ \Rightarrow \ c=1$

$\text {Continuous at}\ \ x=2:$

$2 m+1=\dfrac{8}{2} \ \Rightarrow \ m=\dfrac{3}{2}$

b. $\text{Asymptotes:}$

$\text{As} \ x \rightarrow-\infty, 2^x \rightarrow 0^{+}$

$\text{As} \ x \rightarrow \infty, \dfrac{8}{x} \rightarrow 0^{+}$

$\text{Asymptote at} \ \ y=0.$

$\text{There are no vertical asymptotes.}$

Show Worked Solution

a. $f(x)=\left\{\begin{array}{cl}2^x & \text {for } x<0 \\ m x+c & \text {for } 0 \leq x \leq 2 \\ \frac{8}{x} & \text {for } x>2\end{array}\right.$

$\text {Continuous at}\ \ x=0:$

$2^\circ=m(0)+c \ \Rightarrow \ c=1$

$\text {Continuous at}\ \ x=2:$

$2 m+1=d\dfrac{8}{2} \ \Rightarrow \ m=\dfrac{3}{2}$

b. $\text{Asymptotes:}$

$\text{As} \ x \rightarrow-\infty, 2^x \rightarrow 0^{+}$

$\text{As} \ x \rightarrow \infty, \dfrac{8}{x} \rightarrow 0^{+}$

$\text{Asymptote at} \ \ y=0.$

$\text{There are no vertical asymptotes.}$

Functions, 2ADV EQ-Bank 24

The temperature $T$ (in °C) in a greenhouse follows the pattern:

\begin{align*}
T(h)= \begin{cases}10+2 h, & \text {for }\ 0 \leqslant h<6 \\ 22, & \text {for }\ 6 \leqslant h \leqslant 18 \\ 58-2 h, & \text {for }\ 18<h \leqslant 24\end{cases}
\end{align*}

where $h$ is the number of hours after midnight.

Sketch the graph of $T(h)$ for $0 \leqslant h \leqslant 24$ (2 marks)

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At what time(s) during the day is the temperature exactly 18 °C? (2 marks)

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b. $\text{Temperature is 18° at 4 am and 8 pm.}$

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b. $\text{Temperature}=18^{\circ} \ \text{twice (see graph)}$

$10+2 h=18 \ \Rightarrow \ h=4$

$58-2 h=18 \ \Rightarrow \ h=20$

$\therefore \ \text{Temperature is 18° at 4 am and 8 pm.}$

Functions, 2ADV EQ-Bank 21

The function $g(x)=\left\{\begin{array}{ll}a x+b, & \text {for } x \leq 2 \\ x^2-1, & \text {for } x>2\end{array}\right.$

$g(x)$ is continuous at $x =2$ and passes through the point $(0,-5)$.

Find the values of $a$ and $b$ (2 marks)

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Evaluate $g(3)-g(-1)$ (1 mark)

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a. $a=4, b=-5$

b. $17$

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a. $\text{Since} \ f(x) \ \text{is continuous at} \ \ x=2:$

$a(2)+b$	$=2^2-1$
$2 a+b$	$=3\ \ldots\ (1)$

$\text{Since} \ g(x) \ \text{passes through}\ (0,-5):$

$a(0)+b=-5 \ \ \Rightarrow\ \ b=-5$

$\text{Substitute}\ \ b=-5 \ \ \text{into (1):}$

$2 a-5=3 \ \ \Rightarrow\ \ a=4$

b.	$g(3)-g(-1)$	$=\left(3^2-1\right)-[4(-1)-5]$
		$=8+9$
		$=17$

Functions, 2ADV EQ-Bank 4 MC

A parking garage charges according to the following piecewise function

$C(t)= \begin{cases}5, & \text{for }\ 0<t \leq 1 \\ 5+3(t-1), & \text{for }\ 1<t \leq 4 \\ 18, & \text{for }\ t>4\end{cases}$

where $C$ is the cost in dollars and $t$ is time in hours.

How much does it cost to park for 3.5 hours?

$11.50
$12.50
$13.50
$18.00

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$B$

Show Worked Solution

$C(3.5)=5+3(3.5-1)=\$ 12.50$

$\Rightarrow B$

Functions, 2ADV EQ-Bank 3 MC

For the function $f(x)=\left\{\begin{array}{ll}|x+2|, & \text{for } x \leq 0 \\ \sqrt{x}, & \text{for } x>0\end{array}\right.$, what is $f (-2)+ f (4)$ ?

$2$
$4$
$6$
$8$

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$A$

Show Worked Solution

$f(x)= \begin{cases}|x+2|, & \text { for } x \leq 0 \\ \sqrt{x}, & \text { for } x>0\end{cases}$

$f(-2)=|-2+2|=0$

$f(4)=\sqrt{4}=2$

$f(-2)+f(4)=2$

$\Rightarrow A$

\(a(2)+b\)	\(=2^2-1\)
\(2 a+b\)	\(=3\ \ldots\ (1)\)

b.	\(g(3)-g(-1)\)	\(=\left(3^2-1\right)-[4(-1)-5]\)
		\(=8+9\)
		\(=17\)