Recent content by zfunk

so is the domain just all real numbers then?

well i know that it can't be negative, as ln only works for values above 0

ahh okay ill try to figure it out, so how would i go about working out k for when fg(x)=k with no solution? a figure that would not be within the domain and range?

i see that it works periodically now but how do you actually translate it into writing? sort of confused, thanks

i got b=-4! thanks wil make sure i do! really appreciate you taking time out to help me here thanks!

would it then be [-pi/2,pi/2]? thanks

i got x=2! thanks! however in the question it says the point P lies on curve V and has coordinates (a,b). So i was trying to sub the b=-2a into the dy/dx equation

sorry don't understand what you mean by periodic in this case. thanks!

i get 6x/2x=3 so 3=3, so does that mean a =1? ahh this is soo confusing!

would the Range be (-infinity,infinity)? and domain (0,pi/2]?

i think i get a=0? but not 100% on that

Yes they are from the other site, so how could you write that in interval form?

however the question asks for a value for a and b. so what would they be?

hi again, its functions f and g have domains (0,infinity) and (0,pi/4] respectively and are f(x) = lnx and g(x)=tanx. write the domain and range of fg(x) (for which i got D(0,pi/4] and R(-infinity,0]) solve fg(x)=-0.4 (for which i got 33.84) and give a value of k so that when fg(x)=k has no...

i get this far but when subbing in b=-2a i don't get a proper answer and can't understand what going wrong