Question

1. sandra took her broken laptop to be repaired. the total repair cost includes a fixed cost of $30 plus $18 per hour for labour.

a) use function notation to write an equation for this situation. define your variables. (1 mark)
b) state a reasonable domain and range for your function. explain your thought process. (2 marks)

1. using your knowledge of transformations and inverse functions, and the parent function $f(x) = \sqrt{x}$, determine the domain and range of the inverse of the function $g(x) = 13\sqrt{-(x + 20)} - 41$. explain your thought process. (3 marks)

Explanation:

Question 7

Step1: Define variables

Let $x$ = labour hours ($x\geq0$), $C(x)$ = total repair cost.

Step2: Build cost function

Fixed cost + hourly labour cost.
$C(x) = 30 + 18x$

Step3: Identify domain bounds

Labour time can't be negative; laptop repairs take at least 0 hours, and realistically, a reasonable maximum is, for example, 10 hours (most laptop repairs don't exceed this). So domain: $0\leq x\leq10$.

Step4: Calculate range bounds

Substitute domain endpoints into $C(x)$.
Minimum cost: $C(0)=30+18(0)=30$
Maximum cost: $C(10)=30+18(10)=210$
So range: $30\leq C(x)\leq210$

Step1: Find domain of $g(x)$

Square root argument must be non-negative.
$-(x+20)\geq0 \implies x+20\leq0 \implies x\leq-20$

Step2: Find range of $g(x)$

Parent function $\sqrt{x}$ has range $y\geq0$. Transformations: vertical stretch by 13, shift down 41.
$13\sqrt{-(x+20)}\geq0 \implies 13\sqrt{-(x+20)}-41\geq-41$
So range of $g(x)$: $y\geq-41$

Step3: Relate inverse domain/range

Domain of inverse = range of original function; Range of inverse = domain of original function.

Answer:

a) Let $x$ represent the number of labour hours (where $x\geq0$), and let $C(x)$ represent the total repair cost in dollars.
$C(x) = 18x + 30$

b) Domain: $0\leq x\leq10$ (where $x$ is labour hours). A laptop repair cannot take negative time, and most standard laptop repairs do not take more than 10 hours.
Range: $30\leq C(x)\leq210$ (in dollars). The minimum cost is the fixed $30 (0 hours of labour), and the maximum cost for 10 hours of labour is $210, which aligns with the reasonable domain.

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Question 8