if f(x)=2x-3 and g(x)=2x^2+1 find
A.f(g(2)) B. g(f(x))
C. does f(x) have and inverse if so find the inverse of f(x)

Question

anonymous · Answer

f surely has an inverse because it is a line and therefore one to one.

anonymous · Answer

f says 'multiply by 2 then subtract 3' so f inverse would say to do the opposite things in the opposite order: add 3 and divide by 2 so 
$$f^{-1}(x)=\frac{x+3}{2}$$

anonymous · Answer

if this is confusing, rewrite 
$$f(x)=2x-3$$ as 
$$y = 2x-3$$ then switch x and y (because that is what the inverse does) to get 
$$x=2y-3$$ and solve this for x:
$$x=2y-3$$
$$x+3=2y$$
$$\frac{x+3}{2}=y$$ and y is your inverse.

anonymous · Answer

is that the answer for C

anonymous · Answer

yes. was it clear?

anonymous · Answer

ok so whats the answers for a and b

anonymous · Answer

$$f(g(x))=f(2x^2+1)=2(2x^2+1)-3$$

anonymous · Answer

i guess there is a little algebra to do now:
$$2(2x^2+1)-3=4x^2+2-3=4x^2-1$$

anonymous · Answer

but it was f(g(2))

anonymous · Answer

is it clear what i did?
first write $$f(g(x))$$ then replace $$g(x)$$ by $$2x^2+1$$ and then rewrite f replacing 
$$x$$ by $$2x^2+1$$

anonymous · Answer

oh well if it is $$f(g(2))$$ then since $$f(g(x))=4x^2-1$$ then $$f(g(2))=4(2^2)+1=17$$

anonymous · Answer

typo sorry.
$$4(2^2)-1=15$$

anonymous · Answer

or you could say $$g(2)=2(2^2)+1=9$$
and $$f(9)=2\times 9 - 1=15$$

anonymous · Answer

ok

anonymous · Answer

$$g(f(x))=g(2x-3)=2(2x-3)^2+1$$

anonymous · Answer

this requires more algebra:
$$2(2x-3)^2+1=2(2x-3)(2x-3)+1=2(4x^2-6x+9)+1$$
$$=8x^2-12x+18+1=8x^2-12x+19$$ if my algebra is correct.

anonymous · Answer

ok for a how come its 4(2^2)-1

anonymous · Answer

and not 2(2^2) +1