f(x) =(1-x)/(x^2-2)
if x = -1
f(-1) = 2/(-1) = -2
so no, (-1 ,-2) is on the graph
bottom is zero if x = sqrt 2 or if x = -sqrt 2 so domain is all real x except those two points
f(1/2) = (1/2)/(1/4 - 2)
= (2/4)/(-7/4)
= -2/7
Let f(x)=(x-1)/(-x^2+2)
a. Is this point (-1,-2/3) on the graph of f?
b. State the domain of f and then solve f(x) 1/2?
3 answers
Damon,
For a. you just plug in -1.
f(-1)=(-1-1)/((-1)^2+2)= -2/3 ?
so yes,(-1,-2/3) on the graph.
For a. you just plug in -1.
f(-1)=(-1-1)/((-1)^2+2)= -2/3 ?
so yes,(-1,-2/3) on the graph.
if x = -1
- x^2
is
-(-1*-1)
-(1)
-1
but
(-x)^2
(-1)^2
is
+1
- x^2
is
-(-1*-1)
-(1)
-1
but
(-x)^2
(-1)^2
is
+1