HElp me ASAP?

Question

An incandescent light bulb costs $0.65 to buy nad $0.004/h for electrecity run it. A fluorescent bulb costs $3.50 to buyand $0.001/h to run.
a)use function notation to write a cost equation for each type of bulb.
b)state the domain and range of each equation
c)after how long is the fluorescent bulb cheaper than the regular bulb?
d)deetermine the difference in costs after one year. Assume the light is on for an average of 6h/day

Answer 1

a) f(x)=0.65+0.004x for incandescent bulb, and
g(x) = 3.50+0.001x for fluorescent bulb

where x=time in hours and bot f and g functions measure cost


b) Domain of each function is [0,infinity), since time can't be negative. Range in this case is what the functions worth at time=0: [0.65,infinity) for f(x) and [3.5,infinity) for g(x)


c) we need to find when f(x)=g(x), so:
0.65+0.004x=3.5+0.001x
0.003x=2.85
x=950 hours or 39 days and 14 hours if lit up the full 24 hours


d) 1 year=365x6=2190 hours
f(2190)=0.65+0.004*2190= 9.41$
g(2190)=3.5+0.001*2190= 5.69$

So difference in cost is: 9.41-5.69=3.72$

Answer 2

Call x is the hr using the bulb

a/ Cost for reg bulb: 0.65 + 0.004x
Cost for Fluo bulb: 3.50 + 0.001x

b/ Domain for both: x>=0 y --> Range (.65,infi) and (3.5,infi)

c/ 3.50 + 0.001x < 0.65 + 0.004x
0.003x > 2.85 --> x > 950 hrs
After 950 hrs

d/ diff cost each yr:
x = 365 * 6 = 2190 hr

3.50 + 0.001 * 2190 - 0.65 - 0.004*2190 = -3.72

Answer 3

a) f(t) = 0.65 + 0.004t for incandescent light
g(t) = 3.50 + 0.001t for fluorescent light

b)domain is all positive numbers, range is >0.65 and >3,5

c) solve 0.65 + 0.004t > 3.5 + 0.001t

d) substitute t = 6x365 in both eq and find the difference