Given: $f (x) = 2 + \sqrt{- 3 x + 1}$ $f(x) = 2+ sqrt(-3x+1)$ , $q (x) = \frac{5 x}{x + 2}$ $q(x) = (5x)/(x+2)$ , and $m (x) = ∣ ∣ 7 - x^{2} ∣ ∣ + x$ $m(x) = abs(7 - x^2) + x$ , how do you find f(-33)?

Question

2557 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-10T10:34:02

You can simply substitute x with -33 in f(x) function:
$f (- 33) = 12$ $f(-33)=12$ .

$f (x) = 2 + \sqrt{- 3 x + 1}$ $f(x)=2+sqrt(−3x+1)$

$q (x) = \frac{5 x}{x + 2}$ $q(x)=(5x)/(x+2)$
$m (x) = ∣ 7 - x^{2} ∣ + x$ $m(x)=∣7−x^2∣+x$

How do you find f(-33)?
$f (- 33) = 2 + \sqrt{- 3 \cdot - 33 + 1} = 2 + \sqrt{100} = 2 + 10 = 12$ $f(-33)=2+sqrt(-3*-33+1)=2+sqrt(100)=2+10=12$

Given: f(x) = 2+ sqrt(-3x+1)f(x)=2+√−3x+1f(x) = 2+ sqrt(-3x+1), q(x) = (5x)/(x+2)q(x)=5xx+2q(x) = (5x)/(x+2), and m(x) = abs(7 - x^2) + xm(x)=∣∣7−x2∣∣+xm(x) = abs(7 - x^2) + x, how do you find f(-33)?