f(x) = rad(64 s =^2)
I get
s <= + - 8
I know by looking at the graph that the domain is > - 8, and less than 8, but I'm not sure how to get there algebraically.
I get
s <= + - 8
I know by looking at the graph that the domain is > - 8, and less than 8, but I'm not sure how to get there algebraically.
-
I am assuming that f(x) = √(64 - x^2)
I believe your problem stems from the fact that square rooting both sides of an inequality is not a valid operation; that is, you cannot go from x^2 ≤ 64 to x ≤ ±8. Instead, we need to apply the methods required to solve non-linear inequalities.
Note that we can write:
x^2 ≤ 64 ==> x^2 - 64 ≤ 0
==> (x + 8)(x - 8) ≤ 0, via difference of squares.
Now, (x + 8)(x - 8) = 0 when x = ±8 (by the zero-product property). As for the less-than case: notice that (x + 8)(x - 8) is negative when x + 8 and x - 8 have opposite signs (since a positive number times a negative number equals a negative number and a negative number times a positive number equals a negative number); this occurs when either:
x + 8 ≤ 0 AND x - 8 ≥ 0 ==> x ≤ -8 AND x ≥ 8, which is impossible; OR,
x + 8 ≥ 0 AND x - 8 ≤ 0 ==> x ≥ -8 AND x ≤ 8 ==> -8 ≤ x ≤ 8.
Therefore, the domain of f(x) is [-8, 8].
I hope this helps!
I believe your problem stems from the fact that square rooting both sides of an inequality is not a valid operation; that is, you cannot go from x^2 ≤ 64 to x ≤ ±8. Instead, we need to apply the methods required to solve non-linear inequalities.
Note that we can write:
x^2 ≤ 64 ==> x^2 - 64 ≤ 0
==> (x + 8)(x - 8) ≤ 0, via difference of squares.
Now, (x + 8)(x - 8) = 0 when x = ±8 (by the zero-product property). As for the less-than case: notice that (x + 8)(x - 8) is negative when x + 8 and x - 8 have opposite signs (since a positive number times a negative number equals a negative number and a negative number times a positive number equals a negative number); this occurs when either:
x + 8 ≤ 0 AND x - 8 ≥ 0 ==> x ≤ -8 AND x ≥ 8, which is impossible; OR,
x + 8 ≥ 0 AND x - 8 ≤ 0 ==> x ≥ -8 AND x ≤ 8 ==> -8 ≤ x ≤ 8.
Therefore, the domain of f(x) is [-8, 8].
I hope this helps!