Question

If f(x) is decreasing for all x then

Answer 1

\[\large f(x)=\frac{asinx+bcosx}{csinx+dcosx}\]

Answer 2

1) ad-bc>0
2)ad-bc<0
3)ab-cd>0
4)ab-cd<0

Answer 3

Take the first derivative, find out what pair of inequalities insure its negative for all x

Answer 4

\[\large f'(x)=\frac{( csinx+dcosx)(acosx-bsinx)-(asinx+bcosx)(ccosx-dsinx)}{(csinx+dcosx)^2}\]

Answer 5

\[\large (csinx+dcosx)(acosx-bsinx)<(asinx+bcosx)(ccosx-dsinx)\]

Answer 6

what now?

Answer 7

I am to lazy to do the work, can I ask why you care? What is this homework?

Answer 8

I suppose you could just go with the flow and foil out the stuff...
see what you get...

Answer 9

no idea how to simplify further..

Answer 10

what did you get after you foil out?

Answer 11

answer is B let me know if u can get it smhw

Answer 12

$$f(x)=\frac{a\sin x+b\cos x}{c\sin x+d\cos x}\\f'(x)=\frac{(a\cos x-b\sin x)(c\sin x+d\cos x)-(a\sin x+b\cos x)(c\cos x-d\sin x)}{(c\sin x+d\cos x)^2}$$Note our denominator is always positive so shift attention to the numerator:$$ac\cos x\sin x+ad\cos^2 x-bc\sin^2x-bd\sin x\cos x\\-ac\sin x\cos x+ad\sin^2 x-bc\cos^2 x+bd\cos x\sin x\\=ad(\sin^2 x+\cos^2 x)-bc(\sin^2x+\cos^2x)\\=ad-bc$$

Answer 13

clearly when \(ad-bc<0\) our derivative is negative and our function is always decreasing

Answer 14

such patience.... and attention span -.-

Answer 15

lol,good good :| i was scared to FOIL it