Graph of 4 Lines

[nycmathguy]

Section 1.8
Question 72

See attachment.

Part (a)

I say f(p) = Line 3.

Part (b)

I say g(p) = Line 4.

Part (c)

I know that (g ° f)(p) means g(f(p)).

I know that f(p) = Line 3.

So, g(f(p)) = g(Line 3).

Part (d)

I know that (f ° g)(p) means f(g(p)).

I know that g(p) = Line 4.

So, f(g(p)) = f(Line 4).

You say?

How would you describe the situation in parts (c) and (d)?

 

[MathLover1]

a)

L2
(0,0) and (10,5)=>slope 1/2=0.5

y-0=(1/2)(x-0)
y=(1/2)x -> (1/2)x is 50% of x (in this case 50% discount)

so L2 is f(p)=(1/2)p


b) L1
(5,0) and (10,5)=>slope 1 and x intercept 5
y-y1=m(x-x1)
y-0=1(x-5)
y=x-5=> -5 represents $5 discount so this is function g(p)
g(p)=p-5
so, L1 is g(p)=p-5

c)
(g o f) (p)=g(f(p))
(g o f) (p)=g((1/2)p)
(g o f) (p)=(1/2)p-5 => L4

d)
(f o g) (p)=f(g(p))
(f o g) (p)=f(p-5)
(f o g) (p)=(1/2)(p-5)
(f o g) (p)=(1/2)p-5/2 => L3

 

[nycmathguy]

a)

L2
(0,0) and (10,5)=>slope 1/2=0.5

y-0=(1/2)(x-0)
y=(1/2)x -> (1/2)x is 50% of x (in this case 50% discount)

so L2 is f(p)=(1/2)p


b) L1
(5,0) and (10,5)=>slope 1 and x intercept 5
y-y1=m(x-x1)
y-0=1(x-5)
y=x-5=> -5 represents $5 discount so this is function g(p)
g(p)=p-5
so, L1 is g(p)=p-5

c)
(g o f) (p)=g(f(p))
(g o f) (p)=g((1/2)p)
(g o f) (p)=(1/2)p-5 => L4

d)
(f o g) (p)=f(g(p))
(f o g) (p)=f(p-5)
(f o g) (p)=(1/2)(p-5)
(f o g) (p)=(1/2)p-5/2 => L3

I messed up the entire set of questions.