[a] f(x)=1/(1+x^2) where x belongs to the set of real numbers (Use a sketch of the graph y=1+x^2 to help you sketch f(x))
1/something is always less than 1 , so you known that your graph won't come above the y=1 line.
Unless that something is a fraction itself (dividing by a fraction is multiplying with the fraction upside down).
1+x^2 is always more than one , so that solves that problem of dividing with fractions.
Now you imagine the graph for x^2 , add one , and turn it upside down , but remember where it goes to (actually quite quick)
Hint : 1/something between 0 and infinity never reaches zero as well.
f(x)=3+(x-3)^0.5 where x is greater than or equal to 3
so you got 3 , (x-3) and (x-3)^0.5
first of all , everything less than 3 is non existend on the x axis. (3 + something eh ?)
then the x-3 .. We get the info that x>=3 , so it's always 0 or bigger
3+(0 or bigger)
Now we'll need the graph from x^0.5, add a few things up and away the graph goes.
I want to know how can i sketch the graphs without using calculus for the turning points, etc. (particularily for part [a]) Tia if anyone knows
You need to know the basic graphs , such as x^2 , x^0.5 and 1/x and it's easy.