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?
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
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.
if x = -1
- x^2
is
-(-1*-1)
-(1)
-1
but
(-x)^2
(-1)^2
is
+1
To check if a point is on the graph of a function, we need to substitute the x-coordinate of the point into the function and see if it gives us the y-coordinate.
a. To determine if the point (-1, -2/3) is on the graph of f(x), we substitute x = -1 into the function f(x) and check if it equals -2/3.
Substituting x = -1, we have:
f(x) = (x-1)/(-x^2+2)
f(-1) = (-1-1)/(-(-1)^2+2)
= (-2)/(-1+2)
= (-2)/1
= -2
Since f(-1) = -2, the y-coordinate does not match -2/3. Therefore, the point (-1, -2/3) is not on the graph of f.
b. The domain of a function is the set of all possible input values (x-values) for which the function is defined.
In this case, the function f(x) = (x-1)/(-x^2+2) is defined for all real numbers except the ones that make the denominator, -x^2+2, equal to zero.
To find the domain, we set the denominator equal to zero and solve for x:
-x^2 + 2 = 0
Adding x^2 to both sides, we get:
x^2 = 2
Taking the square root of both sides, we have:
x = ±√2
So, the domain of f is all real numbers except x = ±√2.
To solve f(x) = 1/2, we substitute 1/2 for f(x) in the function and solve for x:
1/2 = (x-1)/(-x^2+2)
To simplify the equation, we cross-multiply:
2(x-1) = -x^2 + 2
Expanding the equation and rearranging terms:
2x - 2 = -x^2 + 2
Moving all terms to one side:
x^2 + 2x - 4 = 0
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.