C3 function question

My question
today q .jpg

My answer to part a

\displaystyle f(x)=\frac{ax+b } {cx+d}
f(x)=f(-x)
\displaystyle =\frac{ax+b} {cx+d}
\displaystyle = \frac{-ax+b} {-cx+d}
(ax+b)(-cx+d)=(-ax+b)(cx+d)

2adx = 2bcx
ad = bc
But for the part about f(x) = k
My working was:

\displaystyle =\frac{ax+b} {cx+d}
\displaystyle =\frac{d(ax+b)} {b(cx+d)}
\displaystyle =\frac{adx+bd} {bcx+bd}
\because ad=bc
\displaystyle =\frac{bcx+bd} {bcx+bd} = 1
1 = constant
\therfore f(x)= k
Please can you tell me if my method is right way to show that f(x) = k

b)
my answer:

\displaystyle f(x)=\frac {ax+b} {cx+d}
\displaystyle =\frac {(ax+b)} {(cx+d)}
\displaystyle = \frac{-ax+b} {-cx+d}
(ax+b)(-cx+d)=(-ax+b)(cx+d)

2bd = 2acx^2
db = acx^2
\displaystyle \frac{bd} {ac}=x^2
\displaystyle x = \pm \sqrt{\frac{bd} {ac}}
\displaystyle \because a= d = 0 or b = c = 0
if a = d = 0
x = undefined

\displaystyle \ therefore f(x) = \frac{k} {x}

Please show me a better way, I try to use my result to show that f(x) =k/x but i dont think i am right. Also how to i show that f(x) = kx

c)i did part c

please help with part a and b

thank you

(edited 7 years ago)

Reply 1

B_9710

Original post by bigmansouf

My question
today q .jpg

My answer to part a

\displaystyle f(x)=\frac{ax+b } {cx+d}
f(x)=f(-x)
\displaystyle =\frac{ax+b} {cx+d}
\displaystyle = \frac{-ax+b} {-cx+d}
(ax+b)(-cx+d)=(-ax+b)(cx+d)

2adx = 2bcx
ad = bc

But for the part about f(x) = k
My working was:

\displaystyle =\frac{ax+b} {cx+d}
\displaystyle =\frac{d(ax+b)} {b(cx+d)}
\displaystyle =\frac{adx+bd} {bcx+bd}
\because ad=bc
\displaystyle =\frac{bcx+bd} {bcx+bd} = 1
1 = constant
\therfore f(x)= k
Please can you tell me if my method is right way to show that f(x) = k

b)
my answer:

\displaystyle f(x)=\frac {ax+b} {cx+d}
\displaystyle =\frac {(ax+b)} {(cx+d)}
\displaystyle = \frac{-ax+b} {-cx+d}
(ax+b)(-cx+d)=(-ax+b)(cx+d)

2bd = 2acx^2
db = acx^2
\displaystyle \frac{bd} {ac}=x^2
\displaystyle x = \pm \sqrt{\frac{bd} {ac}}
\displaystyle \because a= d = 0 or b = c = 0
if a = d = 0
x = undefined

\displaystyle \ therefore f(x) = \frac{k} {x}

Please show me a better way, I try to use my result to show that f(x) =k/x but i dont think i am right. Also how to i show that f(x) = kx

c)i did part c

please help with part a and b

thank you

For the first part you have

f(x)=\frac{d(ax+b)}{d(cx+d)}= \frac{adx+bd}{cdx+d^2}

. You know that

ad=bc

. Using this you should be able to take out a factor on top and bottom containing an expression in x.

(edited 7 years ago)

Reply 2

bigmansouf

Original post by B_9710

For the first part you have

f(x)=\frac{d(ax+b)}{d(cx+d)}= \frac{adx+bd}{cdx+d^2}

. You know that

ad=bc

. Using this you should be able to take out a factor on top and bottom containing an expression in x.

for part a (ii) f(x0 = k I did what you recommended whereby I found b/d. the Problem is that b/d is not a constant to my understanding sine b/d can be any number doesn't that make it a variable

Reply 3

B_9710

Original post by bigmansouf

for part a (ii) f(x0 = k I did what you recommended whereby I found b/d. the Problem is that b/d is not a constant to my understanding sine b/d can be any number doesn't that make it a variable