Question

If mc021-1.jpg and mc021-2.jpg, what is the range of mc021-3.jpg?

Answer 1

$u(x)=-2x^2+3$, and $v(x)= \frac{1}{x}$, 

we want to find the range of $(u\circ v)(x)=u(v(x))$


Notice that Whatever value v can possibly produce, that is the Range of v, becomes the range of u.

Then whatever u produces for these values, is the range of  u(v(x))


So, first find range of v: clearly the domain of v is R-{0}, as v(0) makes no sense.

We check what values c can v produce:

$\frac{1}{x}=c\\\\x= \frac{1}{c}$,

this means that any c (for now) can be produced... it is enough to let x=1/c

this also means that c cannot be equal to 0 as 1/c makes no sense if c=0.


Thus the range of v is R-{0}, 


Now we check the range of u(v(x)) for v(x)∈R-{0}, 

assume we want to produce a value c, so: 

$c=-2x^2+3\\\\-2x^2=c-3\\\\x^2= \frac{c-3}{-2}$

since the left side is always positive or 0, for x=0, the right hand side expression must also be positive or zero, which means c-3 must be negative or 0, 

thus c-3≤0;  c≤3.  Here recall that x in u(x) cannot be 0, so c<3.



Answer: The range of u(v(x))   is (-infinity, 3)